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REESE  LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA 


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•Accession  No.  / 0 Jj    /^  •   Class  No. 


PRACTICAL   NOTES 


— ON— 


HYDRAULIC    MINING. 


—BY— 


GEO.  H.  EVANS,  M.  E. 


SECOND   E 


OF  THB 

UNIVERSITY 


San  Francis 

JOHN  TAYLOR  &  Co., 

63  First  Street 

Vtf   ~\ 
•  i     >-* 


INDEX 

To  Practical  Notes  on  Hydraulics,  by  G*  H.  Evans. 


A. 

Air  Valves . .  .19  j* 

Angles,  loss  of  head 27,  28  7 

Area  of  Ditches 10, 13, 14 

B. 

Banks  of  Ditches 13  I 

Beams,  strength  of. 34,  39  « 

Bends,  loss  of  head 27,  28  « 

Breast  Wheels 32,  33  2 

Bull  Wheels 29  ? 

Bursting  Strain  of  Pipes,  Plates,  etc 42,  43  d 

C. 

Capacity  of  Pipes  and  Nozzles , 14  ^ 

Chains,  strength  of 39,  40  £ 

Construction  of  Nozzles 15  J 

Creeks,  measuring  supply 4  . 

"        to  measure  supply 4  , 

Current  Wheels 30,  31  J 

D. 

Dams,  pressure  of  water 15  * 

Diameter  of  Valves 22  * 

Discharge  of  Pipes 36  ,J 

Discharge  through  Gates 20,  21  J 

Discharge  through  Sluice  Valves 21  £ 

Ditch  By- Washes 7  I 

Ditch  Sidelings 7  ' 

Ditches,  allowance  for  leakage 11  4 

area  of 10,13,14  -J 

"        capacity  of 9  * 

discharge  of 10, 11  -J 

' '         friction  in  same 8  * 

gradeof. 9,10 

"         height  of  sides. 8  ; 

"        in  porous  ground 9  ^ 

"         laying  out  line  of 7  ' 

"         springs  on  line  of 7 

"        timber  work  on 7 

"         tunnelling  through  spurs,  etc 7 

"        weak  banks 13 

*'         wetted  border 10  M 


Elevator,   Hydraulic 44,  47 

Evaporation,  loss  due  to ., 4,11 

Expansion  Joints 19 

F. 

Fall  of  Pine  Line 15 

Flumes,  discharge  of 11 

"  grade  of 8 

height  of  sides 8 

"  iron  or  steel 7 

Friction  Head  for  Pipes 16 

Friction,  in  pipes 24,  25,  26,  27 

G. 

Gates,  careless  opening  and  shutting 20 

discharge   through .20,  21 

head   required 21 

"       screws   of , 19 

Grades,   correct 5 

"        nature  of  country 6 

of  ditches 9 

of   sluices 1 47,  48 

Gravel  Elevators 44,  47 

H. 

Head,  due  to  velocity 19 

loss  due  to  bends 27,28 

required  for  gates  and  valves 21 

Hydraulic  Elevator 44,  47 

Motors 28 

I. 

Iron  or  Steel  Flumes 7 

Iron  and  Steel,  strength  of 40,  41 

J. 
Joints  of  Pipe 20 

L. 

Laying  Pipe  Line 15 

Leakage,  loss  due  to 4,  11 

Loss  due  to  Evaporation  and  Leakage 4 

M. 

Materials,  strength  of 33,  34,  42 

Methods  of  Measuring  Water  Supply 4,  5 

Motors,    Hydraulic 28 

N. 

Nozzles  22,  23,  24 

capacity   of '. 14 

construction  of 15 

V 

O. 

Overshot  Wheels,..  ...33 


Perimeter  or  Wetted  Border 10 

Pipe  Air  Valves 19 

Pipe  Line,  fall  of 15 

"       "       laying  of 15 

Pipes,  bell-mouthed  entrance 16 

bursting    strain 42,  43 

capacity   of 14 

discharge  of 16 

discharge  through  gates 20,  21 

expansion  joints 19 

friction   head 16 

friction  in 24,  25,  26,  27 

joints  of 20 

loss  due  to  bends 27,  28 

plates,  etc.,  bursting  strain 42,43 

safety  valves 19 

velocity  of  water  in 16,  17,  18,  19 

Plates,  iron  and  steel,  strength  of 42,  43 

R. 

Reservoirs,  pressure  of  water 15 

Ropes,  strength  of 40,  41,  42 

S. 

Safety  Valves 19 

Screws  of  Gates 19 

Sizes  of  Sluices 47,  48 

Sluices,  grade  of 47,  48 

narrow  vs.  deep 47,48 

size  of 47,  48 

Sluice  Valves,  discharge  and  head  required 21 

Strength  of  Beams 34,  39 

of  Chains 39,  40 

of  Iron  and  Steel 40,  41 

of  Ropes 41,  42 

Supply,  average  of  water 4 

T. 

Timber  Work  on  Ditches 7 

Trestle  Work,  foundation  for 7 

Tunnels,  form  of  roof 8 

;;A,.    u. 

Undershot  Wheels 31,  32 

V. 

Valves,  diameter  of 22 

"       sluice,  discharge  of 21 

Velocity,  head 19 

"        mean    5 

of  water  through  pipes 16,  17,18,  19 

to  find  in  creeks,  ditches,  etc , 5 

W. 

Water,  average  supply  of 4 

Elevator   44,  47 

Facilities  4 

pressure   of •  •  •  •  •  •  •  J|> 

Wheels  28,  29,  30 

Wetted    Border 10 

Wheels,  breast 32,  33 

bull   30,31 

"         overshot    33 

undershot   ....31,  32 

water  ..  28,29,30 


Practical  Notes  on  Hydraulic  Mining. 


The  following  article  was  written  for  the  MINING  AND  SCIENTIFIC 
PRESS  by  GEORGE  H.  EVANS,  C.  E.,  M.  E.,  General  Manager  Cons. 
G.  Mines  of  Gal.,  Ld.,  Oroville,  Cal. 


COPYRIGHTED. 


Of  the  various  kinds  of  mining  there  are  none  more 
interesting  than  hydraulic  mining,  and  in  connection 
with  it  there  are  innumerable  points  on  which  a  mine 
manager  or  superintendent  should  be  thoroughly 
posted,  some  of  the  most  important  being  as  follows: 

First — Water  facilities  and  the  different  methods 
of  both  roughly  and  accurately  measuring  the  average 
amount  of  water  available  the  season  through  for 
working  the  claim  or  claims. 

Second — The  nature  of  country  through  which  the 
ditches,  flumes  and  pipe  lines  have  to  be  constructed 
in  order  to  carry  water  for  mining  purposes,  and  the 
different  grades  suitable  for  such  purpose.  This  is 
most  important.  The  writer  knows  instances  where 
ditches  have  been  constructed  for  long  distances  with 
too  heavy  a  grade,  and  consequently  the  water  when 
turned  in  acquired  too  much  velocity  and  completely 
washed  away  the  greater  portion  of  the  ditch.  On 
the  other  hand,  by  the  employment  of  cheap  or  in- 
competent men,  ditches  have  been  constructed  with 
too  little  grade,  and  there  are  cases  where  men  have 
constructed  ditches  with  the  fall  in  the  wrong  direc- 
tion. 

Third — The  quantity  of  water  different  size  pipes 
will  carry  or  that  can  be  discharged  through  pipes 
and  nozzles  under  various  heads. 

Fourth — The  friction  caused  by  using  pipes  of  too 
small  diameter,  and  the  loss  of  head  due  to  this;  also 


loss  of  head  due  to  bends  of  short  radius,  and  angles 
of  all  kinds  in  pipe  lines. 

Fifth — A  full  and  complete  practical  knowledge  of 
the  different  motors  used  in  connection  with  hydrau- 
lic mining,  and  all  particulars  relative  to  the  efficiency 
of  the  various  kinds  of  water  wheels,  etc. 

Sixth — The  strength  of  materials,  and  especially  of 
chains,  hemp,  manilla  arid  wire  ropes. 

Seventh — The  bursting  and  working  strain  of  iron 
and  steel  pipes  of  different  diameters,  and  the  strength 
of  iron  and  steel  plates,  single  and  double-riveted,  with 
punched  and  drilled  holes. 

Eighth — Methods  of  economically  treating  alluvial 
deposits  in  large  quantities,  when  there  is  not  suffi- 
cient grade  for  ground  sluices,  and  yet  enough  water 
for  piping,  or  in  cases  where,  owing  to  the  debris 
law,  it  is  necessary  to  impound  the  tailings. 

Water  Facilities. — This  is  one  of  the  most  important 
matters  in  connection  with  hydraulic  mining,  and 
great  care  should  be  taken  in  arriving  at  the  aver- 
age quantity  of  water  available  all  the  year  round, 
or  during  the  entire  season,  so  that  sluicing  opera- 
tions can  be  carried  on  continuously.  In  order  to  do 
this  it  is  necessary  to  correctly  measure  the  creeks,  or 
streams,  at  different  points,  to  fairly  approximate  the 
average  supply  of  water  that  can  safely  be  relied  upon 
from  such  source,  not  forgetting  that,  according  to  the 
location,  allowance  must  be  made  for  loss  due  to 
evaporation  and  leakage,  which  in  some  cases  reaches 
as  high  as  20  per  cent. 

A  very  simple  method  of  measuring  the  quantity 
of  water  flowing  in  a  stream  is  as  follows:  Measure 
the  depth  of  water  in  feet,  at  from  six  to  twelve  points 
across  the  stream  at  equal  distances;  do  this  in  two 
or  three  places  along  a  fairly  straight  course;  add  all 
depths  together,  and  divide  the  result  by  the  number 
of  measurements  taken;  this  will  give  the  average 
depth  of  the  stream,  and  such  depth  multiplied  by  the 
average  width  in  feet  will  give  its  cross  section,  or 


area  in  square  feet,  which,  multiplied  by  the  velocity 
of  water  in  feet  per  minute,  will  give  the  number  of 
cubic  feet  flowing  per  minute  in  the  stream. 

To  find  the  velocity,  a  very  simple  way  is  to  step  or 
measure  off  120  feet  along  the  bank,  and  in  order  to 
allow  for  the  surface  of  the  water  flowing  faster  than 
the  bottom  or  sides,  and  thus  obtain  the  mean  velocity, 
call  the  measurement  100  feet,  and  at  the  commence- 
ment of  this  100  feet  throw  into  the  middle  of  the 
stream  several  pieces  of  paper  or  wood  at  intervals,  and 
note  the  time  it  takes  each  one  of  them  to  reach  the  end 
of  the  measured  line;  then  divide  the  total  time  in  min- 
utes taken  by  all  the  floats  by  the  number  of  floats, 
and  the  result  will  be  the  average  time  taken  for  each 
float  to  make  the  trip ;  divide  the  average  time  in  min- 
utes by  the  distance  in  feet,  viz,  100,  and  the  result  will 
be  the  velocity  in  feet  per  minute,  and  this  multiplied 
by  the  area  in  square  feet  will  give  the  number  of 
cubic  feet  flowing  per  minute;  or,  if  the  answer  be 
required  in  miners'  inches,  multiply  the  cubic  feet  per 
minute  by  2  and  divide  by  3. 

Another  simple  method  for  small  streams  is  to  put 
a  small  dam  across  the  stream  and  back  up  the  water 
sufficiently  deep  to  prevent  any  considerable  velocity, 
and  on  top  of  the  dam  place  a  thin  board  with  a  notch 
cut  out  of  it  wide  enough,  by  estimation,  to  carry  the 
whole  of  the  water  with  a  moderate  depth  of  over- 
flow, and  the  following  calculation  will  give  the  num- 
ber of  gallons  discharged  per  minute,  and  this  result 
divided  by  11.25  will  convert  the  gallons  per  minute 
to  miners'  inches.  For  example:  A  weir  with  4 
inches  overflowing  the  length  of  a  notch  which  is  6 
feet,  or  72  inches,  wide,  the  number  of  gallons  per 
minute  would  be  found  by  the  following  formula: 
G  =  dXV"dXlX  2.67. 

Where  G  represents  gallons  per  minute,  d  =  depth 
of  overflow  in  inches,  and  1  =  length  of  notch  in 
inches.  In  this  case  G  will  be  found  by  multiplying 
4  by  the  square  root  of  4,  and  by  the  length  of  notch 
or  72  inches,  and  then  by  2.67,  making  the  quantity 


—  5  — 


of  water  in  gallons  per    minute  =  1538,    and    this 
divided  by  11.25  =  136.71  miners'  inches. 

There  are  many  other  and  more  correct  methods  of 
measuring  the  flow  of  water  in  channels  and  streams, 
but  I  have  illustrated  the  two  most  simple,  in  order  that 
any  person  of  ordinary  intelligence  could  easily  deter- 
mine the  quantity  of  water  running  in  open  streams 
without  the  aid  of  difficult  formulae. 

Nature  of  Country  for  Grades,  etc. — The  nature  of 
the  country  through  which  it  is  intended  to  carry 
ditches  or  flumes  must  be  carefully  considered  in  order 
to  establish  the  correct  grade,  upon  which,  of  course, 
depends  the  velocity,  or,  more  plainly  speaking,  the 
destroying  force  of  the  water,  and  in  locating  the  sites 
for  water  races  the  following  points  should  be  care- 
fully considered: 

First — Ascertain  by  careful  aneroid  readings  the 
lowest  point  in  the  stream,  creek  or  other  source  of 
water  supply  that  will  allow  sufficient  grade  for  con- 
veying the  water  to  a  point  suitable  for  working  the 
claim  or  claims,  and,  if  the  maximum  supply  obtain- 
able is  less  than  required  for  advantageous  working, 
then  favorable  sites  must  be  located  for  the  construc- 
tion of  storage  dams  or  reservoirs  capable  of  storing 
sufficient  to  keep  up  the  required  supply. 

Second — Should  the  lowest  point  in  the  stream, 
creek  or  other  source  of  water  supply  contain  more 
than  enough  to  meet  the  requirements  of  the  mine  or 
mines,  in  driest  season,  then  the  locator  should  select 
the  greatest  elevation  that  the  country  through  which 
the  ditch  has  to  be  constructed,  and  the  water  supply 
available  at  driest  season  will  allow,  so  that  the  ditch 
when  completed,  will  command  the  largest  area  of 
mining  ground  with  the  maximum  head  or  pressure. 
This  is  an  important  point,  as  in  many  instances 
ditches  of  considerable  length  have  been  constructed 
and  much  money  wasted  in  the  endeavor  to  com- 
mand large  tracts  of  mining  ground,  and  when  such 
ditches  have  been  completed  it  was  found  that  they 

—  6  — 


tapped  the  source  of  supply  at  such  an  elevation  that 
it  was  impossible,  except  in  the  wet  season,  to  get 
sufficient  water  to  wash  with. 

Third — All  timber  work  along  the  line  of  ditch 
should  be  curtailed  as  much  as  possible,  and  when 
fluming  cannot  be  avoided  the  use  of  iron  or  steel 
should  be  carefully  considered  for  ditches  of  a  per- 
manent nature,  as  in  many  instances  the  first  cost  is 
not  very  much  greater,  but  the  durability  and  the  great 
saving  in  cost  of  maintaining  more  than  compensates 
the  owner  of  the  ditch. 

When  timber  work  is  found  necessary,  care  should 
be  taken  in  securing  the  most  durable  kinds,  and  after 
flumes  and  supports  are  finished  they  should  beithor- 
oughly  coated  with  a  hot  mixture  of  asphaltum,  or 
painted  with  a  good  mineral  paint,  while  all  founda- 
tions for  trestles,  etc.,  should  be  placed  in  such  a  man- 
ner that  they  can  be  easily  removed  and  renewed  at 
all  times. 

Fourth — The  line  of  the  ditch  should  be  carefully 
laid  out  so  that  it  will  be  as  short  as  possible,  with, 
of  course,  due  regard  to  economy,  etc.,  and  in  coming 
around  long  points,  or  in  places  where  sidelings  are 
very  steep  and  composed  of  loose  rock,  tunneling 
through  such  spurs  should  be  carefully  considered, 
or  when  it  is  proved  by  boring  that  such  tunnels  can 
be  constructed  to  stand  without  timbering,  they  should 
always  be  preferred  to  long  ditches  around  such  spurs 
or  points,  unless  the  ground  for  ditching  is  exceed- 
ingly good  and  the  extra  distance  quite  short. 

Fifth — Along  the  line  of  ditch  all  springs  or  water 
courses  should  be  connected  by  means  of  short  flumes 
or  ditches,  so  that  the  loss  due  to  leaks  and  evapora- 
tion from  the  main  supply  will  be  entirely  or  partly 
made  up.  It  is  also  absolutely  necessary  for  the  safety 
of  the  ditch  that  by-washes  or  water  gates  be  con- 
structed for  the  purpose  of  taking  care  of  any  sudden 
increase  of  water  from  heavy  rains  or  melting  snow 
along  the  line  of  ditch. 

—  7  — 


These  by-washes  must  be  kept  in  condition  to,  at 
all  times,  divert  any  water  above  the  usual  height  over 
the  gates  or  through  the  openings  in  the  sides  of  the 
race,  and  in  locating  the  points  for  by-washes,  or 
safety  outlets,  it  is  necessary  to  carefully  consider  what 
becomes  of  the  surplus  water,  as  in  many  cases  owners 
of  ditches  have  rendered  themselves  liable  for  heavy 
damages. 

Sixth — At  the  different  points  along  the  line  of  race 
when  fluming  has  to  be  resorted  to,  allowance  should 
be  made  for  an  increase  of  grade,  in  order  that  the 
flume  can  be  constructed  of  much  smaller  dimensions 
than  the  ditch  and  yet  carry  all  the  water  required. 

While  on  this  subject,  it  is  necessary  to  remember 
that  the  least  amount  of  friction  in  ditches  and  flumes 
is  developed  when  the  least  wetted  border,  or  perim- 
eter, is  obtained,  and  to  do  this  the  width  of  the  bot- 
tom must  be  from  If  to  2£  times  the  depth  of  the  sides. 
These  two  points,  if  carefully  studied,  will  save  ditch 
owners  large  sums  of  money  in  both  lumber  and  con- 
struction accounts. 

The  following  is  a  simple  rule  for  finding  the  height 
of  the  sides  of  a  ditch  or  flume  when  area  of  same  is 
known,  and  it  is  desirable  to  follow  the  rule  just  men- 
tioned above : 

When  width  is  to  be  2J  times  the  height  of  the 
sides,  multiply  the  area  in  square  inches  by  4  and 
divide  the  result  by  9,  then  take  the  square  root  of 
the  product  and  that  will  be  the  height  of  the  sides. 

When  the  width  is  to  be  If  times  the  height  of  the 
sides,  multiply  the  area  in  square  inches  by  4  and 
divide  by  7,  then  extract  the  square  root  of  the  product 
and  the  answer  will  be  the  height  of  the  sides. 

Seventh — It  is  agreed  by  the  best  authorities  that, 
when  constructing  tunnels,  where  they  will  stand 
without  timbers,  the  best  form  of  roof  is  the  Gothic 
arch,  as  it  stands  better  than  the  circular  or  any  other 
kind  of  roof  and  is  not  so  liable  to  flake.  In  fact, 
tunnels  constructed  with  circular  roof,  except  in  very 


,^ 

tight  ground,  have  been  noticed  to  flake  off.  until  they 
assume  nearly  the  section  of  the  Gothic  arch. 

Grades,  Capacity,  etc. — In  connection  with  the 
grades  and  various  shapes  of  water  races,  the  follow- 
ing points  require  particular  attention: 

First — As  before  mentioned,  the  character  of  the 
ground  through  which  the  ditch  is  constructed  will 
have  a  great  bearing  on  the  grade  required,  but,  as 
a  guide,  it  will  be  well  to  remember  that  practical 
results  have  demonstrated  that  in  ordinary  ground, 
the  water  should  travel  at  the  rate  of  from  180  to  200 
feet  per  minute.  Then  the  grade  will  be  determined 
by  the  dimensions  of  the  ditch,  and  its  intended  carry- 
ing capacity. 

Second — Races  in  which  the  water  flows  at  too 
high  a  velocity  through  ground  of  a  porous  nature 
will  never  be  free  from  leakage,  owing  to  the  fact 
that  the  velocity  of  the  water  will  not  allow  any  sedi- 
ment to  settle,  and  in  all  ditches  properly  constructed 
the  sediment  traveling  with  the  water  at  a  moderate 
velocity  is  always  relied  upon  to  entirely  tighten  up 
all  portions  of  the  ditch  cut  through  ground  of  a 
porous  nature;  and,  again,  if  the  velocity  is  too  high, 
it  will  scour  holes  in  the  bottom  and  sides  of  the  ditch 
when  constructed  in  sandy  or  clay  soils.  By  neglect- 
ing these  points,  the  cost  of  maintaining  will,  un- 
necessarily, be  increased. 

Third — To  establish  the  grade  of  a  ditch  when  the 
velocity  and  area  of  same  is  known,  one  of  the  sim- 
plest methods  of  calculation  is  as  follows:  Multiply 
the  velocity  in  feet  per  minute  by  the  wetted  perim- 
eter, in  feet,  and  divide  the  result  by  twice  the 
area  in  square  feet,  and  the  product  will  be  the  total 
fall  in  feet  required  to  each  mile.  To  reduce  this  fall 
to  inches,  for  each  12  feet  in  length  multiply  by  .027. 
Example:  Suppose  we  have  a  ditch  to  construct  for 
a  distance  of  six  miles,  to  deliver  600  miners'  inches, 
or  900  cubic  feet,  per  minute,  or  15  cubic  feet  per 
second. 


To  commence  with,  we  are  told  that  in  ordinary 
ground  the  velocity  should  be  about  3  feet  per  sec- 
ond, or  180  feet  per  minute.  Now,  knowing  the 
velocity  and  the  distance,  the  area  required  is  ob- 
tained by  dividing  the  discharge  in  cubic  feet  per 
second,  viz,  15  by  the  velocity  in  feet  per  second,  viz, 
3,  the  result  shows  that  an  area  of  5  square  feet  will 
discharge  the  quantity  of  water  required,  and,  in  order 
to  have  the  ditch  or  flume  constructed  with  least 
amount  of  friction,  the  width  of  the  bottom  must  be 
from  If  to  2%  times  the  height  of  the  sides,  and  in  this 
instance  the  section  of  the  ditch  or  flume  would  be  3 
feet  in  bottom,  with  l-foot-8-inch  sides. 

Having  now  the  velocity  and  area,  we  next  find 
the  wetted  border  or  perimeter — in  other  words,  the 
length — of  so  much  of  the  bottom  and  sides  as  is 
wetted  by  the  water;  for  instance,  if  a  flume  or  ditch 
is  30  inches  wide  and  12  inches  deep,  its  wetted  perim- 
eter, when  full,  ,is  30+12+12=54  inches,  or  4.5 
feet,  and  the  same  ditch  or  flume,  if  empty,  has  no 
wetted  perimeter  at  all.  Now,  fully  understanding 
the  meaning  of  wetted  perimeter,  we  find  that  the 
ditch  or  flume  in  our  example  has  a  wetted  perimeter 
=  to  3  +  1  ft.  8  in.  +  1  ft.  8  in.  =  6  feet  4  inches, 
which,  for  convenience  in  calculating,  we  reduce  to 
decimals,  and  have  6.33.  We  now  have  the  follow- 
ing results,  viz:  Velocity,  3  feet  per  second;  dis- 
charge, 15  cubic  feet  per  second;  wetted  perimeter, 
6.33  feet. 

To  find  the  grade,  we  first  multiply  the  velocity  in 
feet  per  second  by  itself,  and  in  this  instance  the 
result  is  3X3=9,  which  has  to  be  multiplied  by  the 
wetted  perimeter  in  feet,  6.33;  therefore,  9X6.33= 
56.97,  this  total  has  to  be  divided  by  twice  the  area 
in  square  feet,  viz,  5X2=10;  therefore,  56.97—10= 
5.69,  the  total  fall  in  feet  per  mile.  This  result  is 
practically  correct  for  flumes  and  ditches  of  short 
length  in  good  ground,  but  allowance  must  be  made 
according  to  the  roughness  and  the  contour  of  the 
ditch. 

— 10- 


A  more  difficult,  but  correct  formula,  which  has 
been  obtained  from  actual  experiments  made  in  con- 
nection with  ditches  constructed  in  ordinary  ground, 
with  the  usual  winding  course  and  short  bends,  is  as 
follows:  Velocity  in  feet  per  second  =  6  times  the 
square  root  of  2XGXRXS.  Where  G  is  the  accel- 
eration of  gravity,  or  32.2,  R  is  the  hydraulic  radius, 
which  is  found  by  dividing  the  sectional  area  of  the 
ditch  in  feet  by  the  wetted  perimeter  or  border  in 
feet,  and  S,  the  sine  of  inclination,  or  the  total  fall 
or  grade  in  feet,  divided  by  the  total  length  in  feet. 

If  the  ditch  is  constructed  through  rough  country 
and  the  bottom  or  sides  of  same  present  rough  sur- 
faces to  the  water,  then  5  times  the  square  root  of 
2  g  r  s  will  give  the  mean  velocity  in  feet  per  second, 
and  the  velocity  multiplied  by  the  area  in  square 
feet  will  give  the  discharge  in  cubic  feet  per  second, 
which  result  multiplied  by  40  will  give  the  discharge 
in  miners'  inches. 

Example:  To  find  the  velocity  and  then  the  dis- 
charge in  cubic  feet  per  second,  and  miners'  inches 
from  a  ditch  with  a  fairly  straight  course,  and  con- 
structed through  good  ground,  having  the  following 
dimensions  and  fall,  viz,  section  of  ditch,  6X3  feet; 
fall  or  gradient,  8  feet  to  the  mile;  length  of  ditch, 
15  miles. 

We  first  proceed  by  working  out  R,  which  we  are 
told  is  the  sectional  area  of  ditch  in  feet  divided  by 
the  wetted  perimeter  in  feet,  and  in  this   instance 
il '•]      6X3      18 
11 1S  6+3+3=12=  ^ 

S,  or  sine  inclination,  will  be  found  by  dividing  the 

fall  by  the  length,  or  -Jl*  =  .001515. 
uZoO 

Since  twice  G  (the  acceleration  of  gravity)  is  2X 
32.2,  or  64.4,  we  have  G,  R  and  S,  and  our  formula 
stands  as  follows: 

Six  times  the  square  root  of  64.4X1.5X.001515,  and 
the  easiest  method  of  calculation  in  this  case  is  by 
logarithms,  as  follows: 

—  11  — 


Log^s. 

2g=2X32.2  or  64.4=1.8089 

R=1.5=0.1761 

S=.001515=7.1804 

SQUARE  Roox=2  |T9.1654 

"9.5827 

X6         0.7782 

10.3609=2.29 

Answer  2.29  feet  per  second  velocity,  and  this 
multiplied  by  the  area,  18  square  feet=discharge,  or 
41.22  cubic  feet  per  second,  or  41.22X40=1648.80 
miners'  inches. 

The  above  formula  is  also  correct  for  flumes  with 
sawed  boards,  and  battens  over  the  joints  inside  the 
boxes,  but  instead  of  using  6  or  5  as  a  co-efficient 
the  formula  must  read  SV^grs  for  velocity  in  feet 
per  second,  and  8  \X2grs  X  area  for  discharge  in  cubic 
feet  per  second. 

I  might  add  here,  that  these  last  formulae  have 
been  practically  tested  by  several  authorities,  and 
especially  by  the  Government  engineers  of  New  Zea- 
land, to  whom  I  believe  belongs  the  credit  of  arriving 
at  the  exact  co-efficients  shown  above. 

As  before  mentioned,  it  is  of  great  importance  that 
a  safe  allowance  should  be  made  for  loss  due  to  leak-- 
age and  evaporation,  more  especially  when  the  line  of 
ditch  does  not  pick  up  any  small  creeks  or  springs  on 
its  course,  and  it  is  agreed  by  the  best  authorities  that 
a  suitable  allowance  may  be  calculated  by  the  follow- 
ing formula: 

Sectional  area  of  ditch  in  feet 
Mean  velocity  in  feet  per  secondX5280 
Equals  the  loss  in  cubic  feet  per  second  per  mile, 
where  M  is  a  co-efficient  varying  from  3  to  20,  ac- 
cording to  the  climatic  conditions    of    the    country 
through  which  the  ditch  is  constructed. 

In  New  Zealand,  on  the  west  coast,  and,  in  fact,  all 
through  the  middle  island,  good  results  have  been  ob- 

—  12  — 


tained  by  using  3  for  a  multiplier,  but  again  in  the 
North  Island,  where  the  climate  more  resembles  this 
country  and  the  loss  due  to  evaporation  is  heavy,  it 
is  necessary  to  often  use  as  high  as  20  for  M,  in  order 
to  obtain  satisfactory  results. 

Fourth — When  owing  to  weak  banks  it  is  necessary 
to  build  walls  on  the  lower  side  of  a  ditch,  the  ground 
should  be  removed  to  obtain  a  solid  foundation,  and 
two  walls — an  outer  and  an  inner — should  be  built 
up,  with  space  enough  between  to  allow  a  good  pud- 
dle clay  to  be  rammed  in;  such  a  wall,  if  properly 
constructed,  will  never  give  further  trouble. 

Fifth — All  earth,  trees,  roots,  etc.,  must  be  moved 
quite  clear  of  the  lower  side  of  the  ditch,  with  the 
exception  of  just  sufficient  to  make  a  track.  Unless 
all  waste  materials  are  moved  to  such  a  distance  that 
they  will  not  become  a  heavy  drag  on  the  lower  side 
of  the  ditch,  slides  will  be  frequent  and  costly. 

Sixth — At  the  entrance  of  tunnels,  commencement 
of  flumes,  and  at  other  points  where  the  velocity  of 
the  water  is  considerably  retarded,  the  effect  of  water 
changing  its  form  at  such  places  is  an  important  point, 
and  must  never  be  neglected;  in  fact,  all  calculations 
referring  to  the  flow  of  water  in  ditches,  etc.,  the 
mean  velocity  must  be  determined  as  accurately  as 
possible. 

Before  going  further,  many  readers  will  appreciate 
the  following  simple  method  of  arriving  at  the  areas 
or  cross-sections  of  the  different  forms  of  ditches, 
sluices  and  flumes,  which  may  be  calculated  as  follows, 
viz: 

To  find  the  area  of  a  section  of  a  flume  or  ditch 
with  straight  sides,  multiply  the  width  of  bottom  (in 
inches)  by  height  of  sides  (in  inches),  the  product  will 
be  the  area  in  square  inches,  and  this  divided  by  144 
will  give  area  in  square  feet. 

Example :  What  is  the  area  of  a  flume  or  ditch  28 
inches  wide  and  18  inches  deep?  Answer:  28X18=: 
504  sq.  in.,  which,  divided  by  144,  =  3|  sq.  ft. 


—  13  — 


To  find  the  area  of  a  flume  or  ditch  with  sloping 
sides,  add  the  width  at  top  and  bottom  (in  inches) 
together  and  divide  the  result  by  2.  This  answer 
will  be  the  area  in  square  inches,  which,  divided  by 
144,  will  give  the  area  in  square  feet. 

Example :  What  is  the  area  of  a  cross  section  of  a 
ditch  48  inches  wide  at  top  and  26  inches  wide  at  the 
bottom,  with  a  depth  of  24  inches?  Answer:  48  -f- 
26  =  74,  which,  X  24,  =  1776,  and  this  divided  by  2 
=  888  square  inches  ~  144  =  6.16  square  feet. 

To  find  the  square  feet  of  the  cross  section  of  a  ditch 
or  flume  with  sides  sloping  to  a  point  at  the  bottom, 
multiply  the  width  (in  inches)  by  half  the  depth  (in 
inches)  and  the  answer  will  be  area  in  square  inches, 
which,  divided  by  144,  gives  area  in  square  feet. 

Example:  What  is  the  area  of  a  flume  or  ditch  70 
inches  wide  and  sloping  to  a  point  at  the  bottom, 
with  a  depth  of  36  inches?  Answer:  70X18,  or  half 
the  depth  in  inches,  =  1260  inches,  or  divided  by  144 
=  8.75  square  feet. 

Carrying  Capacity  of  Pipes,  Discharge  of  Nozzles, 
etc. — It  is  hardly  possible  to  point  out  any  portions 
of  a  hydraulic  plant  that  are  of  more  importance 
than  pipes  and  nozzles,  and  in  out  of  the  way  places 
miners  have  great  difficulty  in  finding  out  the  cor- 
rect sizes  of  pipes,  and  particularly  the  capacity  of 
same,  especially  with  regard  to  quantities  of  water 
discharged  through  pipes  and  nozzles  of  different 
diameters,  there  being  innumerable  instances  at  the 
present  day  where  miners  do  not  know  the  pressure 
of  water  is  only  as  the  head,  without  any  regard 
(neglecting  friction  and  bends)  to  the  size  of  pipes. 
For  example:  A  pipe  line  composed  of  6-inch  pipe, 
and  another  line  of  40-inch  pipe,  with  same  fall  or 
head,  will  both  give  the  same  pressure.  It  was  only  a 
short  time  since  that  I  was  asked  by  a  miner  of  some 
experience  if  he  could  not  double  his  pressure  by 
doubling  the  diameter  of  his  pipe. 

-  14  - 


As  an  illustration,  this  fact  is  easily  demonstrated 
by  attaching  the  same  size  and  kind  of  faucet  to  two 
tanks,  one,  say,  of  3  feet  diameter  and  3  feet  deep,  and 
the  other  as  large  as  convenient,  say  6  feet  in  diameter 
and  3  feet  deep.  Fill  both  with  water  to  same  depth, 
then,  after  placing  buckets  of  equal  capacity  under 
each  faucet,  turn  both  on  at  the  same  time.  To  the 
surprise  of  any  person  not  acquainted  with  hydraulics 
it  will  be  seen  that,  although  the  larger  tank  contains 
four  times  more  water  than  the  smaller  one,  both 
buckets  will  be  filled  at  the  same  time. 

It  is  well  to  thoroughly  understand  the  principle  of 
hydrostatics  in  building  storage  dams  and  reservoirs, 
remembering  that  there  is  the  same  pressure  on  the 
bank  of  a  reservoir  with  water  3  feet  deep  and  ex- 
tending back  for  a  distance  of  10  feet,  as  there  would 
be  if  the  water  dammed  back  10  miles,  so  long  as  the 
depth  remained  the  same. 

In  reference  to  nozzles,  they  require  great  care  in 
construction  so  as  to  be  of  correct  form,  in  order  that 
the  water  leaving  them  will  be  in  a  solid  stream, 
instead  of  scattering  and  thus  losing  its  power,  which 
is  the  case  with  nozzles  of  improper  construction. 
In  order  to  get  the  best  effect  it  is  absolutely  neces- 
sary that  the  head  of  the  pipe  conveying  the  water 
should  be  at  least  3  or  4  feet  under  water  in  order  to 
prevent  any  air  getting  into  the  pipe,  which  also 
causes  the  water  to  scatter  when  leaving  the  nozzle. 

All  pipe  lines  should  be  laid  as  straight  as  possible, 
or  with  curves  having  as  large  a  radius  as  can  be 
obtained,  and  it  must  be  remembered  that  new  pipes 
well  coated  will  carry  more  water  than  old  pipes 
that  have  been  rusted  inside;  therefore,  allowance 
must  be  made  accordingly.  It  is  good  practice  to 
allow  one-sixth  the  diameter  of  pipes  under  6  inches, 
and  1  inch  on  all  diameters  over  6  inches. 

The  first  thing  necessary  for  the  miner  to  do  is  to 
ascertain  the  fall  available  for  the  pipe  line  and  its 
length  in  feet.  Knowing  this  and  the  quantity  of 
water  he  is  going  to  use  it  is  easy  to  determine  the 


diameter  of  the  pipe,  always  bearing  in  mind  that 
in  order  to  secure  efficiency  and  economy  in  con- 
struction water  should  flow  in  the  pipe  at  a  velocity 
of  not  more  than  3  feet  per  second.  A  few  of  the 
simple  methods  of  determining  the  discharge  of  pipes 
are  as  follows: 

First — To  obtain  the  velocity,  first  multiply  the 
diameter  in  feet  by  the  effective  head  in  feet  and  divide 
the  result  by  the  length  of  the  line  in  feet,  then  take 
the  square  root  of  the  product  and  multiply  by  50; 
this  will  give  the  velocity  in  feet  per  second,  and  the 
velocity  multiplied  by  the  area  of  the  pipe  in  square 
feet  will  give  the  quantity  of  water  discharged  in 
cubic  feet  per  second,  which,  multiplied  by  40,  will 
give  the  number  of  miners'  inches. 

Second — To  find  the  velocity  in  feet  per  minute, 
multiply  the  number  of  cubic  feet  of  water  discharged 
per  minute  by  144  and  divide  the  product  by  the  area 
of  the  pipe  in  inches.  For  example:  An  11-inch  pipe 
discharging  150  cubic  feet  of  water  per  minute,  the 
velocity  would  be  150X144-4-95.03  (the  area  of  pipe 
in  inches)  or  227.6  feet  per  minute,  Or  227.6 Xf =151.7 
miners'  inches. 

Third — In  all  cases  the  pipe  will  require  a  funnel  or 
bell-shaped  entrance,  and  an  additional  head  to  put 
the  water  in  train  in  addition  to  correct  dimensions 
for  overcoming  friction,  assuming,  of  course,  the  total 
head  available  be  required.  To  obtain  the  additional 
head  commonly  termed  the  velocity  head,  a  simple 
method  is  to  square  the  velocity  in  feet  per  second 
and  divide  by  64.4  and  then  divide  that  product  by 
0.70.  The  answer  will  be  the  extra  head  in  feet  re- 
quired. 

We  are  told  by  some  authorities  that  in  cases 
where  the  length  of  the  pipe  exceeds  1000  diameters 
the  head  due  to  velocity  and  even  bends  may  be 
neglected,  but  in  practice  I  find  it  better  to  err  on  the 
right  side,  and  in  no  case  neglect  working  out  the 
heads  due  to  those  losses  and  including  them  in  all 
estimates. 

-16- 


Another  simple  and  approximate  method  of  find- 
ing the  velocity  is  by  multiplying  the  number  of 
.miners'  inches  discharged  by  11  and  divide  the 
product  by  three  times  the  square  of  the  diameter  of 
the  pipe.  For  example:  A  10-inch  pipe  discharging 
400  miners'  inches;  the  velocity  will  be  400X11=4400, 
divided  by  three  times  the  square  of  the  diameter,  or 
3X100=300.  Answer,  14.6  feet  per  second.  By 
means  of  the  same  formula  the  number  of  miners' 
inches  discharged  through  a  pipe  of  a  known  diameter 
and  velocity  will  be  found  as  follows:  Multiply  the 
velocity  in  feet  per  second  by  3  and  the  product  by 
the  square  of  the  diameter,  then  divide  by  11  and 
the  result  will  be  the  discharge  in  miners'  inches. 

For  example:  A  pipe  with  a  diameter  of  20  inches, 
discharging  water  at  a  velocity  of  3  feet  per  second, 
the  number  of  miners'  inches  discharged  will  be  as 
follows:  3X3X400  (the  square  of  the  diameter)  or 
3600^-lli=327  miners'  inches. 

Fourth — A  more  complicated  but  accurate  formula 
for  determining  the  velocity  per  second  of  water  in 
pipes  is:  140  times  the  square  root  of  RXS  minus  11 
times  the  cube  root  of  RXS — where  R  is  the  hydraulic 
radius,  which  is  found  by  dividing  the  diameter  of 
pipe  in  feet  by  4,  and  S  the  sine  of  inclination,  which 
is  found  by  dividing  the  total  fall  in  feet  by  length  of 
pipe  line  in  feet. 

This  formula  has  been  tested  in  a  thoroughly  prac- 
tical manner  by  Mr.  Gordon,  the  Government  En- 
gineer in  New  Zealand,  and  he  found  it  could  not  be 
relied  upon  when  calculating  high  velocities,  as  it 
gave  too  great  a  discharge,  but  with  low  velocities  and 
small  diameters  of  pipe,  it  was  deemed  fairly  accurate. 

Fifth — Another  set  of  formulae  given  in  "Practical 
Hydraulics,"  by  Thomas  Box,  are  as  follows: 

Where  d  =  diameter  of  pipe  in  inches.  L  =  length 
in  yards.  H  =  head  in  feet  G  =  gallons  discharged 
per  minute. 

—  17  - 


,~    x.   ^l/5  .    „          T        (3d)*XH 
d  =  I  ?? I    -*- V  k=r 


H      ^  G2 

yfMrMO    V      H^i  r^'4    V      T 

G=| 


L 

Example  i.  Find  the  diameter  of  pipe  required  to 
discharge  300  gallons  per  minute  with  80  feet  head; 
length  of  pipe,  200  yards: 


80 

Log's. 

300  =  2.4771 
2 


X  200  =  2.3010 

7.2552 

—80  ==  1.9031 

5  |  5.3521 

Fifth  root  =  1.0704 

~-3=  0.4771 


.5933  =  3.92  inches  diameter. 

Example  2.  Find  the  number  of  gallons  discharged 
by  a  pipe  10  inches  in  diameter,  900  yards  in  length, 
with  a  head  of  50  feet: 

X.10)5  X  5<KJ 


uy 


900 
Log's. 

3  X  10  =  30  =  1.4771 
5 

Fifth  power  =  7.3855 
X  50  =  1.6990 

9.0845 

-H  900  ==2.9542 
2  |  6.1303 
Square  root  =  3.0651  =  1162  gallons  per  minute. 

—  18  — 


Example  J.     Find  the  head  necessary  to  discharge 
120  gallons  per  minute  through  a  6-inch    pipe    500 
yards  long: 
_  ISO2  X  500 
(3  X  6)5 

Log's. 

120  =  2.0792 
2 

1202  =  4.1584 
X  500  =  2.6990 

Log.     ^ 
3  X  6  =  18  =  1.2553  X  5  =  6.2765 

.5809  =  3.81  feet. 

In  all  these  examples  care  must  be  taken,  and  more 
especially  in  short  pipe  lines,  to  allow  for  loss  of  head 
due  to  velocity  at  entry,  also  for  friction  of  the  water 
against  the  sides  of  the  pipe. 

Example  4.  Given  diameter  of  pipe  in  inches  and 
velocity  in  feet  per  minute  to  find  discharge  in  cubic 
feet  or  gallons  per  minute.  Q=0.32725Xd2XV= 
cubic  feet  per  minute,  and  cubic  feet  per  minute 
multiplied  by  7.48=gallons  per  minute. 

Sixth — In  constructing  a  pipe  line,  care  must  be 
taken  to  place  air  valves  at  all  high  places,  blow-off 
valves  at  all  low  places  in  the  line,  and,  when  the  line 
is  subject  to  extreme  heat  and  cold,  it  is  absolutely 
necessary  to  provide  at  least  one  good  expansion  joint 
each  half  mile. 

Another  excellent  precaution  is  to  place  near  the 
lower  end  of  the  line  a  safety  valve,  either  the  spring 
or  ordinary  lever  kind,  and  set  it  at  a  pressure  slightly 
above  the  maximum  due  to  the  head,  then  when 
gates  are  closed  too  quickly  by  careless  attendants 
the  valve  will  relieve  the  shock,  instead  of  allowing 
the  whole  line  to  be  strained. 

All  gates  should  have  outside  screws,  and  the 
threads  on  them  fine,  so  that  it  will  be  impossible  to 
open  or  shut  them  too  quickly.  The  careless  open- 

—  19  — 


ing  and  shutting  of  gates  has  wrecked  many  a  good 
line. 

It  has  always  been  my  rule  to  place  tell-tale  gauges 
in  the  main  lines  a  few  feet  behind  a  main  gate,  and 
from  such  a  gauge  one  can  easily  see  whether  the 
water  was  turned  on  or  off  slowly,  and  it  should  be  an 
invariable  rule  to  discharge  any  man  that  handles  a 
gate  in  a  careless  manner,  after  once  being  warned 
by  those  in  charge  of  the  works. 

Seventh — With  reference  to  the  numerous  devices 
for  joining  pipe,  I  have  no  hesitation  in  recommend- 
ing that  all  pipes  near  and  around  the  workings  of 
the  claim  should  be  furnished  with  angle  iron  flanges. 
In  this  country  they  are  certainly  the  exception  to 
the  rule,  but,  if  once  used,  no  other  kind  of  connection 
will  be  tolerated. 

It  is  often  necessary  to  find  the  quantity  of  water 
that  will  be  discharged  through  a  sluice  gate  with 
side  walls,  and  the  number  of  gallons  per  minute  that 
will  flow  through  under  certain  heads  may  be  found 
by  the  following  easy  formula: 

In  determining  the  head  of  water  care  must  be 
taken  to  measure  from  the  center  of  the  opening  in 
sluice  gate  to  surface. 

G  =  8.025  X  1/H  X  .6  X  A  X  6.23  X  60. 

(f        G  x          }* 

H=  <  V6.23  X  60  -*-  A J-s-  .6  > 
I  8.025  j 

Where  G  =  gallons  per  minute,  H  =  head  or  depth 
of  water  from  surface  to  center  of  sluice  opening, 
A  =  area  of  opening. 

Example:  How  many  gallons  per  minute  will  be 
discharged  from  a  reservoir  through  a  sluice  gate 
with  side  walls,  when  the  depth  of  water  above  the 
center  of  the  opening  is  7  feet  and  the  opening  is  & 
feet  wide  and  1  foot  high? 

Answer:  G  =  8.025  X  ^HX  .6  X  A  X  6.23  X  60  = 

E.         A 
8.025  X  V?  X  .6  X  3  ft.  X  6.23  X  60  =  14281.785  gal- 

-20- 


Ions  per  minute.     Which  -f-  11.25  =  1269.49  miners' 
inches. 

Again,  if  it  is  required  to  find  the  head  or  height 
of  water  above  center  of  opening  in  a  sluice  gate 
necessary  to  discharge  14,300  gallons  per  minute 
through  an  opening  3  feet  by  1  foot,  we  proceed  as 
follows: 

f-—      1     1 2 

H  -=  {  V6.23  X  60 —  A^-^-  .6  >    and  in  this  instance 
8.025                     ) 
(f    14300 
H  =  <  V6.2,_  

oi 


=  2.642  or  6.96,  or  nearly  7-inch  head. 

In  many  instances  a  sluice  valve  is  used  instead  of 
a  gate,  and  when  the  pipe  attached  to  the  valves  is 
comparatively  short,  say  of  a  length  not  exceeding 
more  than  three  diameters,  the  following  formula 
may  be  used: 

Where  G  =  gallons  per  minute,  H  =  head  in  feet 
or  height  of  water  above  the  center  of  the  valve  open- 
ing, d  =  diameter  of  valve  opening  in  inches. 

Example  I.  How  many  gallons  per  minute  will  a 
sluice  valve  10  inches  in  diameter  discharge  when 
the  height  of  water  is  3  feet  above  the  center  of  valve 
opening? 

Answer  :  G  =  T/H  X  d2  X  10  =  1/Tx  102  =  10  = 
1732  gallons  per  minute,  or  1732  H-  11.25  =  153 
miners'  inches. 

Example  2.  With  same  measurements  find  the  head 
required  to  discharge  1732  gallons  per  minute. 

Answer: 

G      vL_        1732 


—  21  — 


Example  j.  To  find  diameter  of  valve  necessary 
to  discharge  same  quantity  of  water  with  same  head 
as  in  examples  1  and  2. 

Answer: 


inches  diameter. 

In  connection  with  the  last  formula  it  must  be  borne 
in  mind  that  if  the  pipe  leading  to  the  sluice  valve  to 
reservior  is  much  longer  than  three  diameters,  allow- 
ance must  be  made  for  friction,  etc. 

Nozzles.  —  Nozzles  require  great  care  in  construc- 
tion so  as  to  be  of  correct  form,  and  perfectly  smooth 
in  bore,  in  order  that  the  water  leaving  them  will  be 
in  a  solid  stream,  instead  of  scattering  and  thereby 
losing  power,  as  is  the  case  with  nozzles  of  improper 
construction.  In  order  to  get  the  best  effect  from 
any  kind  of  nozzle,  it  is  absolutely  necessary  that  the 
head  of  the  supply  pipe  should  be  3  or  4  feet  under 
water,  to  avoid  air  getting  into  the  pipe  and  causing 
the  water  to  scatter  when  leaving  the  nozzle. 

To  determine  the  velocity  and  discharge  in  cubic 
feet  per  second  for  well-made  nozzles,  either  of  the 
following  simple  methods  may  be  followed: 

First  —  Multiply  the  square  root  of  the  hydrostatic 
or  effective  head  in  feet  by  8.03.  This  will  give  the 
velocity  in  feet  per  second,  and  that  multiplied  by  the 
area  of  the  discharge  end  of  the  nozzle  in  square  feet 
will  give  the  discharge  in  cubic  feet  per  second,  which, 
multiplied  by  40,  will  give  the  answer  in  miners' 
inches.  Example:  Nozzle  4  inches  in  diameter,  dis- 
charging water  under  an  effective  head  of  400  feet, 
find  velocity  and  discharge.  The  square  root  of  the 
head—  or  400  feet—  is  20,  and  20X8.03  equals  the 
velocity  in  feet  per  second,  viz,  160.60.  Area  of 
4-inch  nozzle  in  square  feet  =  .087266,  and  this  multi- 
plied by  160.60=14  cubic  feet  per  second,  or  14X40= 
560  miners'  inches.  The  result  obtained  by  this  rule  is 
nearly  the  theoretical  discharge,  while  for  ordinary 

—  22  — 


practical  results  the  actual  discharge  will  be  from  75 
to  85  per  cent  of  the  answer  obtained  by  this  rule. 

Second — To  find  the  discharge  in  gallons  per  min- 
ute use  the  following  formulae:  G  =  V  h  X«X  .24. 
Where  G  =  gallons  discharged  per  minute,  h  = 
hydrostatic  or  effective  head  on  nozzle  in  feet,  and 
d  =  the  diameter  of  nozzle  in  -Jths  of  an  inch. 

Example:  Find  the  discharge  from  a  nozzle  3 
inches  in  diameter  with  a  head  of  205  feet.  The  square 
root  of  the  head,  viz,  205  is  14.317  and  the  diameter 
in  £ths  of  an  inch  is  3X8=24,  which  squared  is  24X24, 
or  576.  Therefore,  G=14.317X576X.24  equals  1979 
gallons  per  minute,  and  this  divided  by  11.25=176 
miners'  inches. 

For  accurate  results  the  following  two  rules  may  be 
followed : 

First — Discharge  in  cubic  feet  per  second— V 2Xgh 
XaX0.96,  where  g  is  the  acceleration  of  gravity  in  feet 
per  second,  commonly  accepted  as  32.2  h  =  the 
hydrostatic  or  effective  head  in  feet,  and  a  =  the 
area  of  nozzle  discharge  in  square  feet. 

Second — Discharge  in  cubic  feet  per  second  = 
VhXd2Xc  where  h  =  the  effective  head  in  feet,  d  = 
diameter  of  nozzle  in  -Jths  of  an  inch,  and  c  =  a  vari- 
able co-efficient  from  .00064  to  .00066. 

It  is  a  well  known  fact  that  water  issuing  from  a 
nozzle  should,  theoretically,  attain  the  height  of  the 
head.  For  instance,  a  nozzle  with  300  feet  effective 
head  should  throw  a  stream  a  height  of  300  feet,  but 
we  all  know  this  efficiency  cannot  be  reached  in  prac- 
tice, owing  to  the  resistance  of  the  air,  and  other 
causes,  but  the  difference  has  been  found  by  experi- 
ment to  vary  nearly  in  inverse  ratio  to  the  diameter 
of  the  jet,  and  in  "Practical  Hydraulics,"  by  Thos. 
Box,  we  have  a  formula  for  approximately  calculat- 
ing the  loss  of  head  for  each  case,  which  is  as  fol- 
lows: h'  =^-  X  .0125.  Where  H  =  the  effective 
d 

head  on  the  nozzle  in  feet,  h'  =  the  difference  between 

-23- 


the  head  and  the  height  of  discharge  column  from 
nozzle,  d  =  the  diameter  of  the  jet  in  £ths  of  an 
inch.  Mr.  Box  goes  on  to  remark  that  as  a  result  of 
this  rule  each  size  of  nozzle  attains  a  maximum  height 
with  a  certain  head,  and,  when  the  head  is  increased 
beyond  that  point,  the  nozzle  does  not  throw  the 
stream  so  far,  but,  on  the  contrary,  the  efficiency  of 
the  nozzle  greatly  diminishes;  a  good  deal  owing  to 
the  fact  that  an  excessive  head,  or,  more  plainly  speak- 
ing, a  head  out  of  proportion  to  the  diameter  of  the 
nozzle  tends  to  scatter  the  issuing  stream  and  cause  it 
to  meet  with  more  resistance  from  the  air  than  a  jet 
of  solid  water  issuing  with  a  moderate  head. 

Adopting  Mr.  Box's  formula  we  will  work  out  the 
following,  and  see  the  result: 

First  —  At  what  distance  will  a  well-formed  nozzle 
of  2-inch  diameter  throw  a  stream  of  water  having 
an  effective  head  of  200  feet? 


Answer:  V=          X  -0125or4-^  x  .0125  =  31.25. 

ID  lb  ;  ;  .  ,  . 

Therefore,  the  height  the  nozzle  will  throw  is  200  — 
31.25,  or  168.75  feet. 

Second  —  Take  the  same  nozzle,  with  a  head  of  450 
feet,  and  the  loss  will  be 


X  .0125  or  -  X  .0125  =  158.20  feet. 

That  is  to  say,  the  water  will  be  discharged  to  a  height 
of  450—158.20,  or  291.80,  feet,  instead  of  450  feet,  the 
theoretical  height  minus  all  friction  due  to  the  head. 
Friction  in  Pipes.  —  This  is  a  most  important  matter 
to  those  who  are  connected  in  any  way  with  mining 
or  other  enterprises  in  which  water  is  used  under 
pressure,  and  very  few  miners  are  conversant  with 
the  principles  relating  to  friction  of  water  in  pipes, 
etc.  Most  people  connected  with  water  supply  have 
a  knowledge  of  the  fact  that  when  large  quantities 
of  water  are  discharged  from  pipes  of  small  diameter 
the  pressure  is  greatly  reduced,  but  few  know  how  to 
arrive  at  a  correct  method  of  finding  out  the  exact 


loss  due  to  friction.  Were  it  otherwise  there  would 
not  be  in  evidence  so  many  palpable  blunders  in  the 
construction  of  pipe  lines  used  for  hydraulic  mining 
and  other  purposes,  and  in  many  instances  success 
would  be  the  rule  in  place  of  failures,  many  of  which 
are  due  entirely  to  errors  made  in  bringing  the  water 
supply  to  the  claim,  and  laying  down  pipes  of  too 
small  diameter,  thus  reducing  the  effective  head  or 
pressure  (in  instances  I  have  known)  to  less  than  one- 
half  that  available  with  pipe  lines  properly  propor- 
tioned. 

In  a  previous  paragraph  I  stated  that  water  flow- 
ing through  pipes  should  not  exceed  3  feet  per  second, 
or  180  feet  per  minute,  and  if  all  users  of  water  for 
hydraulic  mining  or  power  purposes  had  their  pipes 
of  the  correct  diameter  to  insure  a  velocity  not  ex- 
ceeding that  named  above  there  would  be  very  little 
trouble,  as  both  efficiency  of  water  and  economy  in 
construction  of  pipe  line  would  be  attained. 

It  must  be  understood  that  the  friction  of  water  in 
pipes  increases  as  the  square  of  the  velocity,  and  also 
depends  upon  the  condition  of  the  pipes  —  whether 
they  are  foul  and  rusty,  or  are  new,  or  in  good  condi- 
tion. Even  the  rivet  heads  in  a  pipe  line  of  consider- 
able length  cause  a  good  deal  of  friction,  and,  con- 
sequently, loss  of  head. 

There  are  several  formulse  for  determining  the 
friction  in  pipes,  but  most  all  of  them  are  difficult  and 
too  complex  for  ordinary  miners.  But  Mr.  William 
Cox  has  simplified  Weisbach's  formula,  and  yet  gives 
identical  results.  Besides  it  is  easy  to  work  out.  It  is 
as  follows  : 

H  -  X  (4  X  V\+  5  V  -  2)  when  H  = 


friction  head  in  feet,  d  =  diameter  of  pipe  in  inches, 
L  =  length  of  pipe  in  feet,  and  v  =  velocity  of  water 
in  feet  per  second. 

Example:     What  is  the  loss  in  head  of  a  pipe  line 
discharging  400  miners'  inches  or  600  cubic  feet  per 


minute,  diameter  of  pipe  being  12  inches,  and  length 
of  line  5,000  feet  We  must  first  find  the  velocity  in 
feet  per  second,  and  to  do  this  we  use  a  simple 
formula,  given  in  remarks  on  velocity  of  water 
through  pipes,  in  a  previous  paragraph,  viz,  multiply 
number  of  cubic  feet  of  water  discharged  per  minute 
by  144  and  divide  the  product  by  the  diameter  of  the 
pipe  in  inches.  Therefore,  in  this  case,  velocity  =  600 
X144—113.10  the  area  in  inches,  or  763.9  feet  per 
minute  or  12.73  feet  per  second. 

Now  knowing  the  velocity,  diameter  and  length, 
we  will  find  H,  or  friction  head,  as  follows: 

4V2  +  5V-2> 

500° 


X-3472 


1200d       12  X  1200 
4  X  12.732-f  5  X  12.73  —  2  =  709.86  and  this  X  .3472 
=  246.44  feet  or  friction  head. 

That  is  to  say,  if  we  had  in  this  example  a  fall  of 
500  feet,  and  constructed  a  pipe  line  with  pipes  12 
inches  in  diameter,  having  a  total  length  of  5,000 
feet,  our  actual  head  of  500  feet  would  be  reduced  to 
500—246.44  or  253.56  feet,  or,  putting  it  more  plainly, 
we  would  have  a  pressure  of  110  pounds  to  the  square 
inch  instead  of  217  pounds,  and  this  loss  is  due  entirely 
to  using  pipes  of  too  small  a  diameter. 

Another  formula  I  use,  and  which  gives  clear  re- 
sults, is  as  follows:  H  =2 

2ga 

H  =  loss  of  head  by  friction  in  each  100  feet  of  pipe. 

p  =  the  perimeter,  or  circumference  of  the  pipe  in 
feet. 

1  =  100  feet. 

c  — a  variable  co-efficient  from  .00406  to  .01338, 
according  to  the  nature  of  the  pipe,  and  velocity  of 
water. 

v  =  velocity  of  water  in  feet  per  second. 

g  =  the  acceleration  of  gravity,  or  32.2  feet. 

a  =  the  sectional  area  of  pipe  in  feet. 

—  26  — 


Example :  A  pipe  line  5000  feet  in  length,  of  newly- 
riveted  pipe,  20  inches  in  diameter,  with  a  head  of  650 
feet  between  the  supply  and  discharge  ends,  and  de- 
livering 400  miners'  inches  of  water,  what  is  the  loss 
of  head? 

First  determine  the  velocity,  which  is  4.58  feet  per 
second,  then 

p  1  c  v2 

_  5.235  X  100  X  .00506  X  20.9  _    55.372 

2  X  32.2  X  2.181  ~  140.456  ~ 

.394  feet,  or  .394  feet  loss  for  each  100  feet  of  line, 
and  there  being  5000  feet  of  pipe,  the  loss  will  be 
5000-^100X.394,  19.70  feet,  and  the  actual  pressure 
head  would  in  this  case  be  650  feet,  less  19.70  feet 
frictional  head,  or  630.30  feet. 

It  is  well  to  remember  that  it  makes  no  difference 
whether  the  water  is  flowing  up  hill  or  down,  or 
whether  the  pressure  is  great  or  small,  the  total  fric- 
tion will  be  materially  the  same,  and  that  in  wooden 
pipes  the  friction  is  nearly  double  that  of  iron  or  steel. 

Loss  of  Head  Due  to  Bends  and  Angles. — This  loss  is 
also  an  important  one,  and  in  many  instances  is  great, 
owing  to  the  number  of  sharp  bends  or  changes  in 
the  direction  of  a  line  of  pipe,  carrying  water  for  min- 
ing, or  other  purposes.  In  a  pipe  line  there  should 
be  no  bends  having  a  radius  of  less  than  five 
diameters. 

To  calculate  the  loss  of  head  due  to  the  resistance 
of  a  right  angle  bend,  the  simplest  rule  is  to  obtain 
the  velocity  of  water  flowing  in  feet  per  second  due 
to  the  head,  and  multiply  the  square  of  such  velocity 
by  .0152.  For  example:  What  is  the  loss  of  head 
due  to  the  resistance  of  a  90-degree  elbow,  with  water 
flowing  at  a  velocity  of  15  feet  per  second?  Answer: 
152X.0152,  or  3.42  feet. 

Where  the  radius  of  the  bend  is  greater,  or  more 
than  5  diameters,  the  head  required  to  overcome  the 
resistance  can  be  found  by  multiplying  the  square  of 
the  velocity  in  feet  per  second  by  the  number  of  de- 
grees in  the  angle,  and  dividing  the  product  by  88489. 


For  example:     Velocity  10  feet  per  second,  what  is 
the  resistance  of  a  bend  having  an  angle  of  120°? 


When  the  radius  is  less  than  5  diameters,  the  resist- 
ance would  be  as  per  following  rule:  Mean  velocity 
squared,  divided  by  64.4,  multiplied  by  the  square  of 
half  the  angle  of  deflection,  multiplied  by  2.06  times 
the  4th  power  of  the  same  angle. 

For  fairly  accurate  results,  this  formula  may  be 
simplified  by  multiplying  the  square  of  the  velocity 
in  feet  per  second  by  C,  C  being  equal  to  the  following 
co-efficients  for  the  various  angles,  viz  : 

C  =  .000109  for  angle  of  10  degrees. 

C  =  .  000466  "  "  "  20  " 

C  =  .001134  "  "  "  30  " 

C  =  .  002158  "  "  "  40  " 

C  =  .003634  "  "  "  50  " 

C  =  .  005652  "  "  "  60  " 

C  =  .  008276  "  "  "  70  " 

C  —  .011491  "  "  "  80  " 

C  =  .  015248  "  "  "  90  " 

Hydraulic  Motors,  Water  Wheels,  etc.  —  There  is  no 
power  easier  handled  or  less  complicated  than  water 
power,  and  it  would  take  too  much  space  to  give  in 
detail  a  history  of  all  the  various  forms  of  motors,  but 
I  -will  endeavor  to  describe  the  particulars  and  effi- 
ciency of  the  most  popular  methods. 

The  power  of  a  fall  of  water  is  easily  calculated,  and 

is  found  as  follows:     Multiply  the  number  of  cubic 

feet  per  minute  by  the  weight  per  cubic  foot,  or  62-J 

pounds,  and  the  product  by  the  fall  in  feet,  then  divide 

by  33,000.     For  example:    What  is  the  horse-power 

in  a  body  of  water  equal  to  60  cubic  feet  per  minute, 

or  40  miners'  inches,  having  a  fall  of  200  feet? 

A  60  X  62.5  X  200 

Answer:  33000        - 


which  is  the  total  power  in  the  water,  and  from  this 
result  allowance  must  be  made  for  friction,  etc. 

Another  easy  method  of  calculating  the  power  is 
to  remember  that  one  cubic  foot  of  water  flowing  per 
minute  and  falling  1  foot  is  equal  to  .0016098  horse- 
power, and  that  one  miners'  inch  falling  1  foot  is  equal 
to  .0024147  horse-power.  These  multipliers  will  give 
result  equal  to  about  85  per  cent  of  the  theoretical 
power. 

Example  i.  What  horse-power  can  be  obtained 
from  40  cubic  feet  of  water  per  minute  falling  300 
feet,  using  a  motor  giving  about  85  per  cent  efficiency? 
Answer:^  .0016098X40X300=19.31  H.  P. 

Example  2.  What  horse-power  can  be  obtained 
from  100  miners'  inches  of  water  falling  60  feet,  using 
a  motor  giving  about  85  per  cent  efficiency?  Answer: 
.0024147X100X60=14.48  H.  P. 

The  most  common  forms  of  wheels  used  by  many 
miners  are  current  or  bull  wheels,  undershot  wheels, 
breast  wheels  and  overshot  wheels.  The  efficiency  of 
these  various  wheels  vary  about  as  follows:  Current, 
or  bull  wheels,  20  to  50  per  cent;  undershot  wheels, 
27  to  35  per  cent;  breast  wheels,  45  to  60  per  cent; 
overshot  wheels,  60  to  75  per  cent.  The  next  class 
of  motors  includes  the  various  forms  of  turbines,  of 
which  there  are  numerous  varieties,  many  of  them 
giving  as  high  as  85  to  87  per  cent. 

Approximate  and  simple  rules  for  finding  quantity 
of  water,  height  of  fall  and  horse-power  developed  at 
an  efficiency  of  75  per  cent  are  as  follows : 

Quantity  of  water  in  cubic  feet  per  minute  is  found 
by  multiplying  the  horse-power  by  706  and  dividing 
the  product  by  the  fall  in  feet.  Example :  How  much 
water  is  required  with  200  feet  of  a  fall  to  develop 
10  horse-power,  using  a  motor  giving  75  per  cent 
efficiency?  Answer:  10X706-f-200=35.3  cubic  feet 
per  minute,  or  about  24  miners'  inches. 

To  find  how  much  fall  is  required  to  generate  a 
required  horse-power,  with  a  known  quantity  of  water, 
multiply  the  horse-power  by  706  and  divide  by  the 
quantity  of  cubic  feet  per  minute. 


Example:  Having  a  supply  of  60  cubic  feet  per 
minute,  and  requiring  20  horse-power  from  motor 
giving  75  per  cent  efficiency,  what  fall  is  necessary? 
Answer:  20X706^60=235.3  feet. 

To  find  the  horse-power  in  a  fall  of  water  when  the 
fall  and  quantity  are  known,  multiply  the  number  of 
cubic  feet  per  minute  by  the  height  of  fall  and  divide 
by  706. 

Example:  Having  a  fall  of  100  cubic  feet  per  min- 
ute, and  a  fall  of  75  feet,  what  horse-power  can  be 
obtained  from  a  motor  giving  75  per  cent  efficiency? 
Answer:  100X75-^-706=10.6  H.  P. 

Should  the  motor  be  of  a  class  that  would  give  only 
65  per  cent  efficiency,  such  as  most  overshot  wheels, 
it  is  necessary  to  use  815  as  a  multiplier  instead  of  706. 

Current  or  Bull  Wheels. — To  calculate  the  horse- 
power of  a  current  or  bull  wheel,  it  should  be  under- 
stood that  in  all  such  motors  the  velocity  of  the  periph- 
ery of  the  wheel,  or,  more  plainly  speaking,  the  num- 
ber of  feet  per  second  the  rim  of  the  wheel  is  traveling, 
should  never  vary  much  from  half  the  velocity  of  the 
stream,  or  half  the  velocity  due  to  the  head  of  water. 
In  this  class  of  wheels  the  diameter  is  seldom  less  than 

6  feet,  or  greater  than  16  feet,  and  the  number  of  floats 

7  to  13.     The  inclination  of  floats  from  radial  lines 
should  be  between  20  and  30  degrees,  depth  of  floats 
from  10  to  16  inches,  and  they  should  be  immersed 
for  about  one-half  their  depth. 

The  horse-power  of  this  class  of  motors  is  found  by 
the  following  formula: 

H  =  .0028XVXMXAX  (V— M). 
M  =  VX.55. 

x  =  cosine  of  angle  between  the  floats,  multiplied 
by  the  radius  minus  the  radius,  or  the  distance  below 
a  horizontal  line  produced  from  under  the  extremity 
of  the  vertical  float. 

Where  H  =  horse-power. 

V  =  velocity  of  current  in  feet  per  second. 
M  =  the  mean  velocity  of  the  periphery  of 

the  wheel  in  feet  per  second. 
A  =  the    immersed    area    of   the    floats    in 
•;'***  square  feet.  *P. 

-  so- 


To  obtain  the  angle  between  the  floats,  divide  360 
(the  number  of  degrees  in  a  circle)  by  the  number  of 
floats  on  the  wheel. 

Example:  A  current  or  bull  wheel,  16  feet  in  diam- 
eter, having  12  floats,  each  of  which  are  8  feet  long, 
with  a  maximum  immersion  of  15  inches,  what  is  the 
horse-power  of  the  wheel,  when  the  stream  has  a  ve- 
locity of  7  feet  per  second? 

Answer:  Angle  between  floats  =  360 .-4- 12  or  30 
degrees.  x  =  the  cosine  of  the  angle  of  30  degrees, 
or  . 86603  X  by  the  radius  or  8  feet  =  6.92824  —  the 
radius,  8  feet  =  1.0717  or  12.86  inches.  This  is  the 
distance  that  the  second  float  will  be  above  the  hori- 
zontal line  produced  from  under  the  extreme  edge  of 
the  vertical  float,  thus  showing  that  although  the 
maximum  immersion  of  any  of  the  floats  is  15  inches 
the  adjoining  floats  would  be  12.86  inches  higher,  and 
to  get  at  the  area  of  the  immersed  floats,  the  depth  of 
the  second  float  will  be  15 — 12.86=2.14  inches,  and 
the  area  of  the  three  immersed  floats=15+2.14-j-2.14 
inches,  or  1.606  feet  multiplied  by  length  of  floats,  or 
8  feet=12.84  feet  area,  and  the  velocity  being  7  feet 
per  second,  M  or  the  mean  velocity  of  the  periphery  of 
the  wheel=7X.55  or  3.85  feet  per  second. 

V    M        A  VM 

Now,  H  =  .0028X8X3.85X12.84X  (7— 3.85)  =  3.49 
H.  P. 

Undershot  Wheels. — To  determine  the  horse-power 
of  an  undershot  wheel,  with  a  rim  .velocity  equal  to 
about  one-half,  or  .57  times  the  velocity  due  to  the 
head  of  water,  or  .57XV2gh:  Where  g  is  the  accel- 
eration of  gravity,  commonly  taken  as  32.2,  and  h  is 
the  head  of  water  in  feet  above  the  bottom  of  the 
wheel, 

H  =.  00066  Qh  or  ^  X  0.35. 
1511  X  H 


-  31  - 


Where  h  =  head  of  water. 

Q  =  quantity  of  water  in  cubic    feet    per 

minute. 

W  =  weight  of  water  in  pounds. 
H  ^effective  horse-power. 

Example:  what  horse-power  is  obtainable  from 
an  undershot  wheel,  with  about  35  per  cent  efficiency, 
using  1500  cubic  feet  per  minute,  with  a  head  of  2 
feet?  Answer:  H=.00066X1500X2=1.98  H.  P.,  or 

Q  =  —    — s —    ~  =1496  cubic  feet  per  minute. 

Breast  Wheels. — The  following  calculations  will  ex- 
plain how  to  arrive  at  the  effective  horse-power  of 
breast  wheels: 

Low  breast  wheels : 

H  =.  00104  Q  X  h. 

961  X  H        H_WXh 
^ "        ~"h~     °r'  *       '  "33000" 
High  breast  wheels: 
H  =  .00108  Q  X  h. 
^       928  X  H  WXh        -_n 

-hT-°r'H  =  :^3000 

Where  Q=quantity  of  water  in  cubic  feet  per 

minute. 

h  =  head  of  water  in  feet. 
H  =  effective  horse-power. 
W  =  weight  of  water  in  pounds. 
Example  I.     A  breast  wheel  16  feet  in  diameter, 
using  1500  cubic  feet  of  water  per  minute,  under  an 
8-foot  head,  what  will  be  the  horse-power    of    the 

Q     h 

wheel?    Answer:    H=.00104X1500X8=12.48  H.  P. 
Example  2.     How  much  water  will    be    required 
under  a  10-foot  head  to  generate  25  horse-power  with 
a  breast  wheel  21  feet  in  diameter?    Answer: 
Q  =  961  XH  =  961-21 

minute. 


In  either  of  the  above  cases  if  the  wheel  takes  the 
water  above  one-half  its  diameter,  of  course  the  power 
would  be  increased,  and  calculations  should  be  made 
by  using  the  formulae  given  for  high  breast  wheels. 
For  example :  A  wheel  22  feet  in  diameter,  using  500 
cubic  feet  of  water  per  minute,  under  a  head  of  16 
feet  (that  is  to  say,  the  water  goes  into  the  buckets  at 
a  point  16  feet  from  the  bottom  of  the  wheel),  what 
horse-power  will  the  wheel  give?  Answer: 

H  =  .00108  X  500  X  16  —  8.64  H.  P. ;  and 

Q  =  -  -  =  500  cubic  feet  per  minute. 

Overshot  Wheels. — I  think  I  am  safe  in  stating  that 
this  class  of  wheel  is  more  popular  than  any  other 
class  of  rough-and-ready  water  motors.  To  find  the 
horse-power  multiply  .00123  by  the  quantity  of  water 
in  cubic  feet  per  minute  used  on  the  wheel.  Then 
multiply  the  result  by  the  head  of  water  in  feet. 

To  find  the  quantity  of  water  required  in  cubic  feet 
per  minute  to  generate  a  given  horse-power  with  a 
known  fall,  multiply  the  horse-power  by  815  and 
divide  by  the  fall,  or  head,  of  water  in  feet. 

Example  I.  What  horse-power  can  be  obtained 
from  an  overshot  wheel  12  feet  in  diameter,  using  200 
cubic  feet  per  minute?  Answer:  .00123X200X12= 
2.95  H.  P. 

Example  2.  How  many  cubic  feet  of  water  per  min- 
ute is  required  to  generate  20  horse-power,  using  an 
overshot  wheel  22  feet  in  diameter?  Answer:  20X 
815^-22=741  cubic  feet 

Strength  of  Materials. — All  those  engaged  in  mining 
should  thoroughly  understand  this  important  subject, 
and  especially  that  branch  relating  to  the  breaking 
and  working  strain  of  ropes,  bolts,  chains,  etc. 

The  number  of  accidents  and  fatalities  arising  from 
ignorance  on  this  subject  should  be  sufficient  to  com- 
pel all  owners  of  mining  property  to  insist  upon  super- 

—  33  — 


intendents  and  foremen  being  able  at  all  times  to  pro- 
vide ropes,  bolts,  chains,  etc.,  of  proper  dimensions 
for  the  work  required,  and  in  this  manner  more  work 
would  be  accomplished,  and  with  far  less  risk  and  ex- 
pense, than  is  generally  the  case. 

I  will  first  deal  with  wooden  beams,  and  as  a  guide 
to  further  calculations  the  following  table  compiled 
from  various  authorities  will  be  found  useful.  The 
columns  marked  S,  N,  C,  E  have  the  following  inter- 
pretation : 

6* — Breaking  load  at  center  of  beam  when  sup- 
ported at  both  ends. 

N — Breaking  load  when  placed  at  one  end  and  the 
other  end  fixed. 

C — Safe  load  in  center  when  beam  is  supported  at 
both  ends. 

E- — Safety  load  at  end  of  beam  when  the  other  end 
is  fixed. 

The  following  table  is  based  on  a  factor  of  safety 
of  7;  that  is  to  say,  the  safe  load  shown  in  table  is 
only  one-seventh  of  the  calculated  breaking  load,  and 
even  this  high  factor  of  safety  should  be  increased 
when  using  beams  not  free  from  knots  and  shake.  In 
addition  to  the  factor  of  safety  freedom  from  knots  and 
shake,  it  must  also  be  remembered  that  seasoned  tim- 
ber resists  crushing  much  better  than  green  timber, 
in  many  cases  twice  as  well,  and  the  figures  given  in 
this  table  are  for  good  samples  of  timber;  therefore, 
the  factor  of  safety  (7)  should  be  adhered  to. 

The  following  figures  and  results  are  obtained  from 
experiments  with  small  pieces  of  timber,  and,  there- 
fore, considerable  allowance  must  be  made  for  beams 
that  are  not  of  a  uniform  texture.  I  have  only  men- 
tioned the  various  kinds  of  American  woods  that  are 
in  general  use.  Their  various  breaking  strength  and 
safe  loads  are  given  in  round  numbers,  so  that  in 
working  out  different  problems  calculations  may  be 
made  as  simple  as  possible: 

-34  - 


Breaking  Safe  load 

Weight  in  Ibs.       load  in  los.                       in  Ibs. 

Name  of  Wood.         per.  cu.  ft.  S.          N.                     C.  E. 

Ash 45  590  147.50  84.29  21.07 

Beech,  white 43  440  110  62.86  15.71 

Beech,red 44  570  142.50  81.43  20.36 

Birch, black 45  680  170  97.14  24.29 

Birch,  yellow 44  440  110  62.96  15.74 

Cedar,  white 35  250         62.50  35.71  8.99 

Fir,  black 42  340         85  48.57  12.14 

Hickory 50  700  175  100  25 

Hickory,  bt.  nut 40  480  120  68.57  17.14 

.Larch 35  300         75  42.96  10.71 

Oak,  live 54  621  155.25  88.71  22.18 

Oak.red 53  562  140.50  80.28  20.17 

Oak,  white 49  581  145.25  83  20.75 

Pine,  red 40  509  127.25  72.71  18.18 

Pine,pitch 41  576  144  82.28  20.87 

Pine,  yellow 33  395         98.70  56.43  14.71 

Pine,  white 34  410  102.50  58.57  14.64 

Pine,  Virginian 38  485  121.25  69.28  17.32 

Teak 56  673  168.25  96.14  26.03 

The  above  figures  represent  the  number  of  pounds 
required  to  fracture  the  various  kinds  of  wood  having 
an  area  or  cross  section  of  1  square  inch  by  1  foot  in 
length.  In  finding  the  strength  of  beams  the  follow-; 
ing  proportions  of  strength  must  be  observed: 

VALUES   OF  X. 

(1.)  With  a  beam  fixed  at  one  end  and  loaded  at 
the  other=l. 

(2.)  With  a  beam  fixed  at  one  end  and  the  load 
distributed  uniformly=2. 

(3.)  With  a  beam  supported  at  both  ends  and 
loaded  at  the  center=4. 

(4.)  With  a  beam  firmly  fixed  at  both  ends  and 
loaded  at  the  center=6. 

(5.)  With  a  beam  supported  at  both  ends  and  uni- 
formly loaded=8. 

(6.)  With  a  beam  firmly  fixed  at  both  ends  and 
uniformly  loaded=12. 

In  calculating  the  strength  of  beams,  the  whole 
weight  of  the  material  must  be  included  when  the 
beam  has  a  uniform  load  and  only  half  the  weight  of 
material  when  the  load  is  placed  at  the  center. 

Let  S=tabular  number  of  breaking  load  in  pounds 
on  a  beam  supported  at  both  ends  and  loaded  at  the 
center. 

-  35- 


N~tabular  number  for  breaking  load  in  pounds  on 
a  beam  loaded  at  one  end,  and  the  other  end  firmly 
fixed. 

C=Safe  load  in  pounds  on  center  of  beam  sup- 
ported at  both  ends. 

E=safe  load  in  pounds  on  a  beam  fixed  at  one  end 
and  loaded  at  the  other. 

b  =  breadth  of  beam  in  inches. 

d  =  depth  of  beam  in  inches. 

1  =  length  of  beam  in  feet. 

w  =  breaking  load  in  pounds. 

R  =  w  divided  by  7=safe  load  in  pounds. 

x  =  proportion  of  strength  due  to  position  of  load, 
and  method  of  fixing  the  ends  of  the  beam. 

Square  Beams. — The  formula  for  finding  breaking 
load  is  as  follows: 

bd2N 
w  =  — = —  X  x 

Example  I.     What  is  the  breaking  load  on  a  beam 
of  American  yellow  pine,  12  inches  deep,  12  inches 
broad  and  20  feet  long,  one  end  of  same  being  firmly 
fixed  and  the  load  at  the  other  end? 
Answer: 

b        d2          N 

12  X  122  X  98.7  _  12  X  144  X  98.7         _ 

w=="        ~20~  ~20~ 

1 
85027.68  Ibs.,  or  37.95  tons.     And  the  safe  load  would 

37.95       ,  ,0  ^ 
be  — = —  or  5.42  tons. 

Example  2.  With  same  dimensions,  but  beam  sup- 
ported at  both  ends  and  loaded  in  the  center,  the 
breaking  load  will  be: 
b        d2          N 

12  X  1?nX  98'7  X  *  =  37.95  tons  X  4  =  151.80  tons. 
20  ,  no 

l 


ff  U 

151.80 
And  the  safe  load  would  be  — ^ —  or  21.68  tons. 

Example  j.  If  the  beam  in  example  1  was  uni- 
formly loaded,  the  safe  load  would  be  the  same,  viz, 
5.42  tons  multiplied  by  x,  and  by  referring  to  the 
different  proportions  of  strength  tabulated  we  find  in 
this  instance  where  the  beam  is  fixed  at  one  end  and 
uniformly  loaded,  x=2.  Therefore  the  safe  load  would 
be  5.42X2=10.84  tons,  thus  showing  that  a  beam  will 
safely  stand  double  the  load  when  uniformly  loaded 
than  it  will  with  load  in  the  center. 

Example  4.  A  beam  of  American  pitch  pine,  6 
inches  wide  and  10  inches  deep  by  15  inches  long, 
supported  at  both  ends  and  uniformly  loaded,  the 
breaking  and  safe  loads  will  be  as  follows: 

w  =  — - —  X  x  Safe  load  =  -= 

b       d2         N 

fi  V  102  V  144-  x 

w  =  g_A        *         =  576°  lbs-  X  8  =  4608°  lbs->  or 
\°  20.57  tons. 

20.57 
And  the  safe  load  would  be  — ^ —  or  2.94  tons. 

Example  5.  Take  the  same  beam  as  used  in  ex- 
ample 4  and  lay  it  flatwise,  or,  that  is,  call  it  10  inches 
wide  and  6  inches  deep,  then  the  breaking  strain 
would  be: 

b       d2       N  „ 

10  V  ft2  V  144-  x  IkS. 

L^£     ±  =  3456  X  8  =  27648,  or  12.34  tons. 

1  tons. 

And  the  safe  load  would  be  — '•=—  =  1.76  tons, 

or  a  little  more  than  one-half  the  strength  of  the  same 
beam  laid  edgewise. 

Round  Beams — In  order  to  find  the  strength  of  a 
circular  beam  it  is  necessary  to  first  work  out  the 
breaking  load  of  a  square  beam  of  which  each  side  is 
equal  to  the  diameter  of  the  circular  beam,  and  multi- 

-  37  — 


ply  this  load  by  .589,  so  the  formula  for  the  break- 
ing load  will  read: 

W  =  J*1J?  XXX  0.589. 

Example  I.  Having  a  round  beam  of  American 
cedar,  40  feet  long  between  supports  and  uniformly 
loaded,  with  a  diameter  of  10  inches,  the  breaking  load 
would  be: 


1  7362 

3.29  tons,  and  the  safe  load  would  be  -=-  =  1052  Ibs. 

Oval  Beams  —  In  this  case  first  find  the  load  for  a 
rectangular  or  square  beam,  with  sides  equal  to  the 
two  diameters  of  the  oval  beam,  and  multiply  the 
result  by  0.6. 

Example.  Having  an  oval  Learn  of  American  white 
pine,  firmly  fixed  at  both  ends  and  loaded  in  the 
center,  having  15  feet  between  supports,  the  smallest 
diameter  10  inches,  the  largest  diameter  12  inches,  and 
placed  so  that  it  would  be  10  inches  wide  and  12  inches 
deep,  the  breaking  strain  or  load  would  be: 

bd2N        '    '•*          ' 
---  j-  —  X  x  X  0.6. 

Here,  on  referring  to  the  multiplier  given  in  pro- 
portion of  strength,  we  find  x=6,  and,  therefore, 

w  iboxiWi(L  j  x  0  6  »  35424     or 

15 

1  15  8 

15.8  tons,  and  the  safe  load  would  be  —  ^—  =  2.26  tons. 

Triangular  Beams  —  To  find  the  breaking  or  safe 
load  for  this  class  cf  beams,  first  find  the  strength  of 
a  square  beam  with  equal  sides  and  divide  the  result 
by  3. 

According  to  experiments  made  by  Barlow  and 
others,  it  was  found  that  triangular  beams  were  J- 
stronger  where  the  base  of  the  triangle  was  up,  and 

—  38  — 


it  was  found  necessary  to  provide  in  the  support  a 
triangular  notch  in  which  to  place  the  sharp  edge  of 
the  beam. 

Strength  of  Chains. — The  strength  of  chains  varies, 
owing  to  the  natur-  of  the  iron  from  which  they  are 
made  and  their  mechanical  construction.  The  strength 
also  varies  as  the  square  of  the  diameter  of  the  iron 
from  which  the  links  are  made.  Experiments  show 
that  a  single-link  chain  from  good  iron  carefully 
welded,  made  from  1  inch  diameter  round  bars,  has  a 
safe  working  strain  of  six  tons.  Great  care  must  be 
exercised  in  using  chains  on  loads  that  would  cause 
disaster  in  case  of  breaking,  and  each  link  should  be 
carefully  examined,  always  bearing  in  mind  that  the 
strength  of  the  chain  is  only  equal  to  the  strength  of 
the  weakest  link.  Many  serious  accidents  have  been 
caused  by  not  paying  proper  attention  to  this  fact. 

There  are  many  formulae  for  determining  the  safe 
load  for  and  breaking  strength  of  chains,  all  of  which 
are  only  approximate  and  depend  upon  careful  exam- 
ination prior  to  attaching  the  load.  For  crane  chains 
the  breaking  load  can  be  found  by  multiplying  the 
square  of  the  circumference  of  the  link  in  inches  by 
32 A,  and  for  the  safe  load  divide  the  result  by  6. 

Example:  What  is  the  breaking  load  of  a  chain 
made  with  links  from  a  round  bar  of  iron  f  of  an  inch 
in  diameter?  Answer:  752X32.4=18.225  tons,  and 

,    t      ,      18.225       _.. 
safe  load  = — - — =  3.04  tons,  nearly, 
b 

Another  simple  and  approximate  result  is  to  divide 
the  square  of  the  diameter  of  the  iron  from  which  the 
links  are  made  by  9,  and  the  result  is  the  safe  work- 
ing strain  or  load. 

Example :  A  chain  made  from  round  iron  J  inch  in 
diameter,  the  safe  load  would  be  4  (the  number  of 
eighths  of  an  inch  in  -J  inch)  squared,  or  16  divided 
by  9=1.8  tons,  the  safe  load. 

The  same  result,  nearly,  is  obtained  by  squaring 
the  diameter  of  the  link  iron  in  inches  and  multiplying 
by  7.111.  Taking  the  above  example  the  answer 
would  be:  .52  or  .25X7.111=1.77  tons. 


To  find  the  diameter  of  the  iron  in  eighths  of  an  inch 
that  the  links  should  be  made  from,  to  safely  support 
a  given  load,  proceed  as  follows:  Multiply  the  weight 
to  be  hoisted  or  hauled  in  tons  by  9  and  extract  the 
square  root  of  the  product,  and  the  answer  will  be  the 
number  of  eighths  of  an  inch  there  should  be  in  the 
diameter  of  the  links. 

Example:  What  sized  iron  should  the  links  of  a 
chain  be  made  from  to  safely  support  a  load  of  two 
tons?  Answer:  V2X9=V  18=4.24  eighths  of  an 
inch,  or  a  little  over  £  inch  in  diameter. 

Some  readers  will  no  doubt  recall  instances  to  their 
minds  where  they  have  lifted  much  heavier  loads  with 
such  chains  as  shown  in  the  examples  given.  Al- 
though this  is  often  done  without  any  bad  results, 
nevertheless  these  rules  should  be  followed  whenever 
possible,  and  especially  when  chains  are  used  for  long 
pulls  and  subject  to  heavy  strains,  as  it  is  much  better 
to  err  on  the  safe  side. 

While  dealing  with  the  strength  of  chains,  it  may 
be  of  interest  and  information  to  readers  to  have  the 
following  table  which  I  have  in  my  pocketbook  and 
often  find  of  great  value  in  making  calculations  as  to 
strength  of  iron,  bolts,  bars,  rivets,  etc.: 

Tensile  Strength  and  Shearing  Strain  of  Iron  and  Steel. 

Tensile  Strength      Shearing  Strength 
per  Sectional  In.      per  Sectional  In. 

Cast  Iron 7  9 

Wrought  Iron  rolled  bars 25  20 

Best  Lowmoor  rivets 29  23 

Cast  Steel,  best  quality  for  tools 52  39 

Double  Shear  Steel 40  30 

Cast  Steel  Boiler  Plates 48  36 

Puddled  Steel  Boiler  Plates 42  31^ 

Bessemer  Steel  Boiler  Plates 32  24 

Steel  Bars 45  31Vi 

Cast  Steel  Rivets 49  37 

Wrought  Iron  Plates ,  length  way 22&  18 

Wrought  Iron  Plates,  crosswise 20V£  1654 

It  is  as  well  to  explain,  for  the  benefit  of  some 
readers  that  tensile  strength  of  any  material  is  the 
weight  attached  to  the  end  of  a  bar  that  will  tear  it 
asunder,  and  the  shearing  strength  is  the  weight  or 
pressure  that  will  cut  the  material  through. 

—  40- 


In  calculating  the  strength  of  screw  bolts,  of  course, 
a  proper  allowance  must  be  made  for  the  thread,  and 
an  approximate  allowance  is  to  deduct  A  from  the 
diameter  of  small  bolts,  and  from  ^  to  i  for  large  bolts. 
For  instance,  a  bolt  %  an  inch  in  diameter,  deduct  -J 
of  an  inch  for  thread  and  calculate  the  strength  of  said 
bolt  as  if  it  were  a  f  bolt,  instead  of  a  ^  inch,  and  with 
a  bolt  2  inches  in  diameter,  deduct  £  and  call  it  If, 
when  calculating  its  working  or  safe  load. 

In  order  to  fully  explain  the  use  of  the  above  table 
of  tensile  strengths  and  shearing  strains,  I  give  an 
example  as  follows: 

What  is  the  tensile  strength  of  a  bar  of  iron  2  inches 
in  diameter? 

Here  2  inches  in  diameter  has  an  area  or  cross- 
section—to  3.141  sectional  inches,  and  3.141X25= 

78  52 
78J  tons,  and  the  safe  load  would  be  —^—=11.22  tons. 

Strength  of  Hemp,  Manilla,  Iron  and  Steel  Rope. — 
This  subject  is  a  most  important  one,  and  every  super- 
intendent and  mine  foreman  should  be  thoroughly 
conversant  with  the  mode  of  calculating  the  breaking 
strain,  and,  more  particularly,  the  safe  load  that  ropes 
of  different  material  will  stand,  as  the  lives  of  the  men 
employed,  especially  in  deep  mines,  are  dependent 
entirely  upon  the  safety  of  the  ropes  used  in  hoisting, 
etc.;  in  fact,  men  working  in  mines  have  a  right  to 
demand  that  the  employers  have  a  thorough  knowl- 
edge of  this  subject. 

A  simple  test  for  the  purity  of  manilla  or  sisal  ropes 
is  as  follows:  Take  some  of  the  loose  fiber  and  roll 
it  into  balls  and  burn  them  completely  to  ashes,  and, 
if  the  rope  is  pure  manilla,.  the  ash  will  be  a  dull  gray- 
ish black.  If  the  rope  be  made  from  sisal,  the  ash  will 
be  a  whitish  gray,  and,  if  the  rope  is  made  from  a 
combination  of  manilla  and  sisal,  the  ash  will  be  of  a 
mixed  color. 

For  calculating  the  breaking  strain  of  round  ropes 
of  different  materials,  the  following  table  is  one  of 
several.    Where  B  =  breaking  strain  in  tons,  and  C 
=  circumference  of  rope  in  inches. 
—  41  — 


B  =  C2XO.m  for  hemp  rope. 

B  =  C2X0.2  ordinary  fiber  rope. 

B  =  C2X1.5  iron  wire  rope  ordinary. 

B  =  C2X2.5  steel  wire  rope. 

B  =  C2X2.09  flexible  galvanized  wire  rope. 

B  =  C2X2.60  extra  flexible  galvanized  wire  rope. 

B  =  C2X4.18  plough  steel  rope. 

The  working  or  safe  load  should  be  taken  as  about  J 

~D 

or  y  of  the  breaking  load  B,  or  - 

6. 

The  weight  of  ropes  can  also  be  approximately  cal- 
culated from  the  circumference,  as  follows:  Where 
W—  weight  of  each  100  feet  in  pounds,  and  C=cir- 
cumf  erence  of  rope  in  inches,  as  follows  : 

W=C2X4.16  for  each  100  feet  of  hemp  or  fiber  rope. 

W=C2X14.54  for  each  100  feet  of  iron  or  steel  rope. 

A  splice  weakens  a  rope  about  one-eighth,  and  it 
is  well  to  remember  that  a  three-strand  rope  is  about 
one-fifth  stronger  than  a  four-strand  one  of  the  same 
dimensions. 

The  Bursting  and  Working  Strain  of  Iron  and  Steel 
Pipes,  Plates,  etc.  —  Under  this  head  I  will  deal  only 
with  wrought-iron  and  steel  pipes  as  are  generally 
used  in  connection  with  mining  work,  and  in  arriving 
at  the  safe  working  strain  or  pressure  it  must  be  under- 
stood that  the  following  rules  depend  upon  good 
workmanship,  correct  diameters  and  distance  apart 
in  riveting,  etc.  In  making  wrought-iron  pipe  care 
must  be  taken  to  have  the  plates  rolled  lengthwise, 
as  it  generally  affects  the  strength  of  the  longitudinal 
seams  —  that  is  to  say,  the  plates  should  be  rolled 
across  the  grain,  and  not  with  it. 

The  simplest  method  of  calculating  the  pressure 
that  wrought-iron  and  steel  pipes  will  stand  is  as 
follows  : 


Where  P~safe  working  pressure. 
T  =  tensile  strength  of  plates,  taking  iron  at  48,000 
pounds  to  the  square  inch  and  steel  at  75,000  pounds. 


t  =  thickness  of  plates  in  inches  or  decimals  of  an 
inch. 

R  =  radius  of  pipe  in  inches. 

f  =  proportional  strength  of  plates,  as  follows: 
when  double-riveted=0.7  and  single-riveted=.5. 

c  =  a  co-efficient  or  factor  of  safety  usually  taken 
at  3. 

p  =  pressure  in  pounds  per  square  inch  due  to  head 
of  water. 

Example:  What  is  the  safe  working  pressure  for 
a  36-inch  pipe,  double-riveted  along  the  longitudinal 
seams,  and  made  from  wrought-iron  plates  rolled 
across  the  grain,  and  •$•  of  an  inch  thick? 

Answer :   P  =  ^^  ^  £  I  X  <K7  =  77-7  Ibs. 
lo 

to  square  inch  or  179  feet  pressure  head. 

By  means  of  the  same  formula,  the  thickness  of  plate 
is  easily  found,  that  will  safely  stand  any  given  pres- 
ure;  for  instance: 

-  p  X  R  X  c 
TXf  " 

Example:  Having  a  pressure  of  179  feet,  or  77.7 
pounds  to  the  square  inch,  what  thickness  must  the 
wrought-iron  plates  be  for  making  a  pipe  36  inches 
in  diameter? 

P       radius  c 

A  77.7  X  18  X  3        10K  .     , 

Answer  :     t  =      — ^ —  —  .125  inch. 

48000  X  0.7 

With  reference  to  the  difference  between  the 
strength  of  drilled  and  punched  holes,  it  has  been 
determined  by  experiment  that  the  loss  of  strength 
in  the  metal  between  the  rivet  holes  when  drilled  is 
practically  nothing,  or,  to  give  the  summarized  result, 
the  loss  when  drilled  lengthwise  was  but  1.13  per  cent, 
and  when  drilled  crosswise  the  loss  was  only  0.9  per 
cent,  while,  on  the  other  hand,  experiments  made 
by  Mr.  Kirkaldy  on  punched  plates  showed  the  mean 
loss  to  be  13  per  cent  with  the  grain  and  17.26  across 
the  grain.  Although  the  result  of  experiments  has 

—  43  — 


conclusively  proved  that  the  strength  of  plates  is 
greater  with  drilled  holes  than  with  punched  holes, 
the  extra  cost  occasioned  by  drilling'  would  not  make 
up  for  the  extra  strength  obtained,  except  in  particu- 
lar cases;  and  with  regard  to  wrought-iron  and  steel 
pipes  for  mining  purposes,  the  effect  of  drilling  or 
punching  need  not  be  taken  into  consideration,  but 
at  the  same  time  I  thought  it  would  be  interesting 
to  some  readers  to  learn  the  difference  in  strength 
between  the  two  methods. 

Methods  of  Treating  Alluvial  Deposits  in  Large  Quan- 
tities, Where  Sufficient  Fall  is  Not  Available,  Sluices, 
Grades,  etc. — Under  this  heading,  I  will  first  deal 
with  what  is  known  as  the  "Hydraulic  Elevator/' 
which  is  the  simplest  and  most  economical  machine 
connected  with  mining  alluvial  deposits,  where,  by 
lack  of  grade,  it  is  impossible  to  run  bedrock  sluices, 
or  when  it  is  impossible  to  secure  a  good  dump  at  the 
end  of  sluices  to  keep  them  running  in  a  proper  man- 
ner. The  elevator  also  enables  large  areas  of  ground 
to  be  profitably  mined  in  districts  where  the  debris 
law  is  operative,  by  means  of  lifting  the  material  into 
restraining  dams,  etc. 

In  this  country,  with  such  water  facilities,  there 
are  thousands  of  acres  that  can  be  profitably  worked 
by  means  of  an  elevator;  but,  owing  to  the  heavy 
cost,  weight  and  poor  efficiency  of  the  elevators  used 
in  California,  it  appears  to  be  regarded  by  miners  that, 
to  work  an  elevator,  it  is  necessary  to  have  excessive 
heads  and  small  lifts,  or,  in  other  words,  one  must 
have  a  large  quantity  of  water,  or  it  is  of  no  use  trying 
to  mine  with  an  elevator. 

Now,  this  is  practically  correct,  when  the  clumsy 
and  costly  machines  that  have  been  the  custom  for 
years  past  are  considered.  It  is  far  from  correct  when 
considering  the  hydraulic  elevator  now  in  the  market 
and  manufactured  in  San  Francisco,  for  which  pat- 
ents have  been  granted  to  the  writer. 

I  do  not  in  any  way  wish  my  readers  to  think  my 
remarks  here  are  for  the  purpose  of  advertising  these 
elevators,  for  my  personal  benefit,  and,  to  be  frank 
with  all,  I  may  state  that  in  making  arrangements  for 

~  44  -^ 


the  manufacture  of  these  machines  I  agreed  upon  a 
price  to  be  charged  that  only  carries  an  ordinary 
manufacturing  profit,  and  the  amount  of  royalty  re- 
ceived on  each  machine  is  but  $75,  my  aim  being  to 
assist  brother  miners,  and,  with  that  end  in  view,  did 
not  insist  on  such  exorbitant  royalties  as  has  been  the 
custom  in  connection  with  other  elevators.  I  feel  a 
little  diffident  about  pushing  forward  the  many  claims 
in  regard  to  cost,  weight,  efficiency,  etc.,  and  would 
much  rather  any  one  interested  would  write  for  par- 
ticulars from  the  manufacturers  in  San  Francisco — the 
Risdon  Iron  Works. 

What  I  wish  to  impress  upon  readers  is  that  if  they 
have  a  piece  of  ground  that  is  too  low  for  ordinary 
mining  and  have  only  a  few  inches  of  water  under 
pressure,  do  not  think  it  is  impossible  to  work  it  by 
means  of  an  elevator.  I  had  charge  of  a  large 
hydraulic  claim  in  New  Zealand,  where  with  less  than 
400  inches  of  water  under  a  working  head  of  225  feet 
I  lifted  sand  and  gravel  to  a  height  of  52  feet,  and 
each  twenty-four  hours  handled  from  2000  to  2400 
tons,  and  for  one  year  elevated  at  the  rate  of  an  acre 
each  month  to  a  depth  varying  from  30  to  35  feet 
banks.  Of  the  quantity  of  water  used,  only  250  inches 
was  taken  by  the  elevator  and  the  remainder  by  the 
giant  for  piping.  The  elevator  was  connected  upon 
the  surface,  and  instead  of  going  to  great  expense, 
as  is  usual  in  this  country,  to  sink  shafts  to  bedrock 
before  placing  the  elevator,  we  let  the  elevator  do  its 
own  sinking,  and  in  less  than  a  week  the  machine 
had  excavated  its  own  shaft  to  bedrock,  a  depth  of 
43  feet  through  some  bad  running  ground,  which  is 
much  better  than  sinking  by  hand,  timbering  through 
running  sand,  let  alone  the  expense  of  pumping,  etc. 

It  is  well  known  by  those  who  have  gone  into  the 
efficiency  of  the  hydraulic  elevator  that,  unless  the 
various  parts  are  well  proportioned,  it  is  a  most  waste- 
ful machine;  but,  as  against  this,  the  enormous 
amount  of  work  they  are  capable  of  doing  with  a 
minimum  cost  for  repairs  and  attendance,  make  them 
especially  valuable  for  hydraulic  mining,  and,  when 

—  45  — 


constructed  on  proper  lines,  there  is  no  difficulty  in 
obtaining  from  40  to  50  per  cent  efficiency  from  same. 

In  this  country  there  are  many  instances  where  10 
per  cent,  and  even  less,  is  only  obtained,  which  is 
partly  due  to  the  machine  being  wrongly  constructed 
and  a  good  deal  to  the  fact  that  most  of  those  at  pres- 
ent using  elevators,  with  few  exceptions,  do  not  know 
how  to  get  the  best  work  from  them. 

There  are  cases  where  superintendents  of  elevator 
claims  appear  to  think  that,  as  long  as  their  machines 
can  be  made  to  handle  boulders,  in  some  cases  20 
and  30  inches  in  diameter,  that  is  a  certain  test  of  their 
efficiency.  Such  is  not  the  case,  as  there  is  no  diffi- 
culty in  lifting  stones  double  or  treble  the  size,  pro- 
viding plenty  of  water  and  pressure  be  available. 

To  be  successful  in  elevating  auriferous  deposits, 
unless  exceedingly  rich,  one  must  make  up  their 
mind  to  use  the  smallest  quantity  of  water,  in  both 
elevating  jet  and  giants,  so  that  the  proportion  of 
gravel  or  other  auriferous  material  lifted  in  compari- 
son with  the  water  required  for  lifting  and  piping, 
will  be  of  a  maximum  quantity.  If  this  rule  was  fol- 
lowed by  many  of  the  present  elevator  claims,  they 
would  be  lifting  and  treating,  in  many  cases,  three 
and  four  times  the  quantity  of  auriferous  material  they 
are  now  doing. 

There  are  cases,  of  course,  where  it  may  be  neces- 
sary, for  a  short  time,  in  opening  a  claim  to  elevate 
the  largest  stones  for  the  purpose  of  making  room, 
but  to  continue  such  course  is  a  .great  mistake — far 
better  stack  the  larger  stones  in  the  paddock  and  use 
the  extra  quantity  of  water  required  for  lifting  such 
stones,  in  working  a  second  or  third  elevator,  and,  in 
such  manner,  be  treating  three  and  four  times  more 
pay  dirt,  than  working  one  heavy  and  wasteful  ma- 
chine to  lift  large  boulders,  etc. 

Let  any  reader  of  this  paper  carefully  work  out  the 
power  there  is  in  the  water  used  through  their  elevator 
jet,  then  find  out  the  power  required  to  lift  the  weight 
of  water  used  for  elevating,  plus  the  weight  of  the 
minimum  quantity  of  necessary  water  for  piping,  and 


—  46  — 


plus  the  weight  of  gravel  or  material  sluiced  and 
lifted  by  elevator;  it  will  then  be  seen  at  a  glance 
whether  you  are  getting  the  proper  work  from  the 
machine  or  not 

I  know  of  an  instance  where  a  manager  of  an  ele- 
vating claim  in  this  State  kicked  because  he  was  not 
getting  75  per  cent  from  his  head  of  water,  when 
using  a  special  form  of  water  wheel,  while  he  was 
using  an  elevator  not  yielding  him  12  per  cent  in  the 
same  claim;  it  never  occurred  to  him  to  try  the  effi- 
ciency of  his  elevator,  as,  from  the  appearance  of  the 
huge  quantity  of  water  and  the  large  and  small  rocks 
coming  through  the  elevator,  he  was  satisfied  the 
efficiency  could  not  be  better.  In  some  respects  this 
was  true,  but  the  owners  of  the  claim  were  paying 
him  to  lift  auriferous  material,  not  water  and  large 
stones,  occasionally  mixed  with  a  few  tons  of  pay  dirt. 

These  machines  are  made  in  convenient  pieces  for 
mule  transportation,  so  that  where  it  is  possible  to 
take  a  mule  there  will  be  no  difficulty  in  erecting  and 
working  an  elevator. 

Grades  and  Sizes  of  Sluices. — There  seems  to  be  a 
great  diversity  of  opinion,  especially  among  the  older 
miners,  as  to  the  size  of  sluices.  Many  of  them  stick 
to  the  idea  that  the  best  results  are  obtained  from 
a  narrow  or  deep  form  of  sluice,  and  will  try  and  make 
others  believe  that  such  a  sluice  is  most  easily  kept 
in  good  running  order.  Now,  this  is  not  correct  by 
any  means,  the  reverse  being  the  case,  it  having  been 
conclusively  proven  by  practical  results  that  a  wide 
sluice  will  carry  more  material,  and  save  more  gold, 
with  considerable  less  grade,  than  a  narrow  and  deep 
sluice. 

It  is  a  well-known  fact  in  relation  to  gold  saving 
that  one  of  the  first  principles  is  to  have  as  thin  a 
film  of  water  and  material  in  the  sluices  as  is  possible, 
with,  of  course,  due  regard  to  the  fact  that  there  must 
be  a  sufficient  depth  of  water  to  just  cover  the  largest 
stones  that  are  sent  down  the  gold-saving  sluices.  I 
have  noticed  on  more  than  one  occasion  when  using 

—  47  — 


narrow  sluices  that  it  takes  much  less  time  to  cause  a 
block  than  in  wide  ones,  and  when  I  state  that  I  have 
worked  sluices  through  ground  containing  nearly  10 
per  cent  of  titaniferous  iron,  or  black  sand,  a  good 
many  miners  will  realize  the  difficulty  in  getting  rid 
of  from  2000  to  2400  tons  of  such  a  deposit  per  day, 
on  a  grade  of  2  inches  to  12  feet,  and  in  doing  this  I 
soon  found  out  the  fallacy  of  using  narrow  and  deep 
sluices. 

With  reference  to  the  grade  of  sluices,  experiments 
made  in  river  gravel  have  shown  that,  with  a  grade 
of  from  1  in  20  to  1  in  25,  40  miners'  inches,  or  60 
cubic  feet,  per  minute  will  wash  from  about  140  to  170 
cubic  yards  per  day  of  24  hours,  and  with  light  grades 
the  depth  of  water  in  sluices  must  be  as  shallow  as 
possible,  just  so  that  it  will  move  the  largest  stones 
and  prevent  the  sand  from  packing.  Mr.  Gordon 
says  that  when  there  is  a  large  proportion  of  heavy 
stones  the  best  results  are  obtained  by  having  about 
from  10  to  12  inches  of  depth  of  water  in  the  sluice, 
this,  of  course,  assuming  that  the  large  stones  do  not 
run  over  10  to  12  inches  in  diameter. 

In  concluding  this  paper  it  is  hardly  necessary  to 
remind  the  reader  that  the  different  formulae  are  not 
original;  but  in  several  instances  they  have  been  sim- 
plified to  meet  the  result  of  my  own  practical  experi- 
ence. The  information  and  data,  on  the  whole,  are 
compiled  from  figures,  formulae,  data  and  notes  ac- 
cumulated in  my  private  pocketbook,  and  if  what  has 
been  written  proves  of  the  same  assistance  to  any  of 
my  fellow  miners  or  to  those  interested  in  mining  as 
myself,  I  will  feel  amply  repaid  for  the  time  taken  in 
preparing  this  paper. 

X'-:- 

IT  rV 


-  48  - 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


2L)eC'59CR| 


LD  21A-50w-4,'59 
(A1724slO)476B 


General  Library 

University  of  California 

Berkeley 


VT       *<*,"> 

I A  0223 


